Chemistry · Chapter 3

Composition of Substances and Solutions

Atoms are far too small to count one at a time. A single grain of salt holds more of them than you could tally in a lifetime. So chemists count by weighing instead. This chapter hands you the tool that makes that trade work. It is called the mole, and it ties the mass on a balance to the number of particles you cannot see. From there you will find the mass of any formula, split a compound into the share of mass each element owns, and work backward from lab data to a formula. At the end you will mix and water down solutions to an exact strength.

By the end of this chapter you can

Use Avogadro's number to move between moles and particle counts.
Find the molar mass of any formula, parentheses and hydrates included.
Convert grams to moles and moles to grams in either direction.
Work out the percent by mass of every element in a compound.
Turn mass or percent data into an empirical formula.
Scale an empirical formula up to a molecular formula.
Mix a solution to a target molarity and dilute a stock solution to order.

Section 1The mole and Avogadro's number

Chemistry needs a counting word

You already use counting words every day. A dozen means 12. A ream of paper means 500. Each word stands for a fixed count, whatever you point it at. A dozen eggs and a dozen bricks hold the same number of things. Their masses are nothing alike.

Chemistry needs a counting word too, but a huge one. A teaspoon of water holds roughly 1.7 × 10²³ molecules. Counting them one per second would take longer than the age of the universe. The mole is the counting word that makes such numbers workable. One mole is 6.022 × 10²³ of whatever you are counting. The unit is written mol.

That count has a name of its own. Avogadro's number, written N_A, is 6.022 × 10²³ particles per mole. The exact value is now fixed by definition. One mole is 6.02214076 × 10²³ particles, no more and no less. Four digits is plenty for the work in this book.

Why such a strange count

The number looks random. It was chosen with great care. It was set so that one mole of carbon-12 atoms weighs almost exactly 12 grams. That link is the whole point of the mole. It means the mass of one mole in grams matches the mass of one particle in atomic mass units. Carbon has an atomic mass of 12.011 amu, so one mole of carbon atoms weighs 12.011 g. You weigh grams on a balance. You get a count for free.

The word particle is doing real work in that definition. A mole counts whatever you aim it at, so you have to say what. A mole of neon is 6.022 × 10²³ atoms. A mole of water is that many molecules. An ionic solid like NaCl has no molecules at all, so we count formula units instead. One formula unit of NaCl is one Na⁺ paired with one Cl^-.

Crossing the bridge

Moles sit in the middle of every problem in this chapter. Mass sits on one side. Particle count sits on the other. To go from moles to particles, multiply by 6.022 × 10²³ per mole. To come back, divide by it. Write the unit on every number and let the units cancel. If what is left over is not the unit you wanted, your setup is wrong, not your calculator.

Figure 3.1 The mole bridge. Molar mass connects grams to moles. Avogadro's number connects moles to a count. There is no shortcut from grams straight to particles.

Worked example · moles to particles and back

You have 2.50 mol of water. How many molecules is that? How many hydrogen atoms?

Step 1 · read the amountYou hold 2.50 mol of H₂O. Three significant figures.

Step 2 · pick the factorOne mole holds 6.022 × 10²³ molecules, so the factor is (6.022 × 10²³ molecules / 1 mol).

Step 3 · multiply2.50 mol × 6.022 × 10²³ molecules/mol = 1.5055 × 10²⁴, which rounds to 1.51 × 10²⁴ molecules.

Step 4 · check the unitsMoles cancel and molecules remain. Three figures went in, so three come out.

Step 5 · go one step furtherEach H₂O carries 2 H atoms. So 2.50 mol H₂O holds 2 × 2.50 = 5.00 mol H. That is 5.00 mol × 6.022 × 10²³/mol = 3.01 × 10²⁴ H atoms.

Step 6 · run it backwardA different sample holds 1.20 × 10²⁴ iron atoms. Divide by Avogadro's number. (1.20 × 10²⁴) / (6.022 × 10²³/mol) = 1.99 mol Fe.

Key points

One mole is 6.022 × 10²³ particles. The unit of N_A is per mole.
A mole counts atoms, molecules or formula units. Always name which one.
Moles to particles, multiply by N_A. Particles to moles, divide by N_A.
Equal moles of two substances hold equal counts. They do not weigh the same.
The mole was built so that one mole of carbon-12 weighs 12 g.

Interactive Walk mass, moles and particles across the mole bridge. Open full ↑

Section 2Molar mass

The mass of one mole

Molar mass is the mass of one mole of a substance. Its unit is grams per mole, written g/mol. For a single element you can read it straight off the periodic table. Take the atomic mass and attach g/mol to it. Carbon is 12.011 g/mol. Oxygen is 15.999 g/mol. Iron is 55.845 g/mol.

For a compound you add up the atoms in the formula. Each subscript tells you how many of that atom to count. Multiply each atomic mass by its subscript, then add. The total is the formula mass in amu. It is also the molar mass in g/mol. The number is the same and only the unit changes.

Water is H₂O. Two hydrogens at 1.008 plus one oxygen at 15.999 gives 18.015. So water is 18.02 g/mol. Through this book we round molar masses to two decimal places. That is already finer than any balance you will use.

Parentheses and raised dots

Some formulas carry a group inside parentheses. The subscript outside multiplies everything within the group. Ca(NO₃)₂ holds one Ca, two N and six O. Miss that and your answer lands far from the truth. The same rule holds for a long formula like Al₂(SO₄)₃, which holds 2 Al, 3 S and 12 O for a total of 342.13 g/mol.

Other formulas carry a raised dot, as in CuSO₄·5H₂O. A solid like this is a hydrate. Water molecules sit inside the crystal in a fixed ratio. The dot does not mean multiply. It means that much water is locked in. Add the water in as five whole units of H₂O. The pentahydrate comes out at 249.68 g/mol. Drive the water off with heat and the dry salt is only 159.61 g/mol. Forgetting the water throws a weighing off by more than a third.

One factor, two faces

Molar mass is a conversion factor you can flip. Written as g/mol it turns moles into grams. Flipped to mol/g it turns grams into moles. Choose the face that cancels the unit you are holding. That choice is the only decision in most of these problems.

Figure 3.2 Molar mass is a sum, not a single lookup. In calcium nitrate the two nitrogens and six oxygens together outweigh the calcium by three to one.

Worked example · molar mass of calcium nitrate

Step 1 · count the atomsCa(NO₃)₂ gives 1 Ca, 2 N and 6 O. The 2 outside multiplies both the N and the three O inside.

Step 2 · weigh each elementCa: 1 × 40.08 = 40.08 g/mol. N: 2 × 14.007 = 28.014 g/mol. O: 6 × 15.999 = 95.994 g/mol.

Step 3 · add them40.08 + 28.014 + 95.994 = 164.088 g/mol.

Step 4 · roundThe molar mass is 164.09 g/mol.

Worked example · grams and moles, both directions

First, grams to moles. You weigh out 25.0 g of glucose, C₆H₁₂O₆.

Step 1 · build the molar massC: 6 × 12.011 = 72.066. H: 12 × 1.008 = 12.096. O: 6 × 15.999 = 95.994. Sum = 180.156, so 180.16 g/mol.

Step 2 · flip the factorYou hold grams and want moles, so use (1 mol / 180.16 g).

Step 3 · divide25.0 g × (1 mol / 180.16 g) = 0.1388 mol, which rounds to 0.139 mol.

Step 4 · now go the other wayHow much does 0.350 mol of NaCl weigh? NaCl is 22.990 + 35.45 = 58.44 g/mol.

Step 5 · multiply0.350 mol × 58.44 g/mol = 20.454 g, which rounds to 20.5 g. Moles cancel and grams are left.

Key points

Molar mass is grams per mole. It matches the formula mass in amu.
Multiply each atomic mass by its subscript, then add every term.
A subscript outside parentheses multiplies the whole group inside.
A raised dot means water locked in the crystal. Count it in.
g/mol turns moles into grams. mol/g turns grams into moles.

Interactive Build a formula and watch its molar mass add up atom by atom. Open full ↑

Section 3Percent composition

Who owns the mass

A pure compound has a fixed recipe. Every sample of pure glucose carries the same share of carbon by mass, whether it came from a grape or a beaker. Percent composition reports that share for each element in the formula. It answers a very practical question. In 100 g of this substance, how many grams are carbon?

The method is one line long. Take exactly one mole of the compound. For each element, divide the mass that element contributes by the molar mass of the whole thing. Multiply by 100 to turn the fraction into a percent. The result for one element is its mass percent.

mass percent of X = (mass of X in one mole ÷ molar mass) × 100%

Notice that the compound's identity is doing all the work. You never need to know how big your sample is. A grain and a truckload give the same percents. That is why percent composition is printed on labels and quoted in catalogs.

A number you can reuse

Percent composition is more than a tidy fact. Once you have it, you can jump straight from the mass of a sample to the mass of one element inside it. Turn the percent into a decimal and multiply. A fertilizer bag marked 35.00% nitrogen says just that. A 50.0 kg bag holds 0.3500 × 50.0 kg = 17.5 kg of nitrogen. No moles needed.

The shares always add to 100%, give or take a little rounding. That hands you a free check. If your percents sum to 94% or 108%, something is wrong. Go back and recount the atoms in your molar mass. A missed oxygen or a dropped parenthesis is the usual culprit.

Figure 3.3 Glucose has twice as many hydrogen atoms as carbon atoms, but hydrogen owns under 7% of the mass. Percent composition tracks mass, never number.

Worked example · percent composition of glucose

Find the mass percent of each element in C₆H₁₂O₆, then use it on a 5.00 g sample.

Step 1 · molar mass72.066 + 12.096 + 95.994 = 180.156 g/mol, or 180.16 g/mol.

Step 2 · mass from each elementCarbon 72.066 g, hydrogen 12.096 g, oxygen 95.994 g, all per mole.

Step 3 · divide and scaleC: (72.066 / 180.156) × 100 = 40.00%. H: (12.096 / 180.156) × 100 = 6.71%. O: (95.994 / 180.156) × 100 = 53.28%.

Step 4 · check the sum40.00 + 6.71 + 53.28 = 99.99%. That is 100% apart from rounding, so the work holds.

Step 5 · use itA 5.00 g sample holds 0.4000 × 5.00 g = 2.00 g of carbon.

Worked example · nitrogen in a fertilizer salt

Ammonium nitrate, NH₄NO₃, is sold for its nitrogen. What percent of its mass is nitrogen?

Step 1 · count carefullyThere are two N, four H and three O. Both nitrogens count, one in the ammonium and one in the nitrate.

Step 2 · molar massN: 2 × 14.007 = 28.014. H: 4 × 1.008 = 4.032. O: 3 × 15.999 = 47.997. Sum = 80.043, so 80.04 g/mol.

Step 3 · divide and scale(28.014 / 80.043) × 100 = 35.00% N.

Step 4 · checkH works out to 5.04% and O to 59.96%. The three add to 100.00%.

Key points

Mass percent is the element's mass in one mole divided by the molar mass, times 100.
Percent composition does not depend on sample size. It belongs to the compound.
Count every atom of an element, even when it appears in two places in the formula.
The percents should sum to 100%. Use that as a built-in check.
Percent as a decimal times sample mass gives the mass of that element.

Interactive See what fraction of a compound's mass each element owns. Open full ↑

Section 4Empirical and molecular formulas

Two kinds of formula

A molecular formula tells you exactly how many atoms sit in one molecule. An empirical formula gives the smallest whole-number ratio of those same atoms. Glucose is C₆H₁₂O₆. Divide all three subscripts by 6 and you get CH₂O. That is the empirical formula of glucose.

Very different compounds can share one empirical formula. Formaldehyde really is CH₂O. Acetic acid is C₂H₄O₂, which also reduces to CH₂O. All three have the very same percent composition. So percent data by itself can never hand you a molecular formula. You need one more measurement, and that measurement is the molar mass.

Ionic compounds are a special case. A crystal has no molecules, so its formula is already the simplest ratio. NaCl and Fe₂O₃ are empirical formulas.

From lab data to a ratio

Analysis gives you masses or percents. It never hands you subscripts. The fix is to turn mass into moles, because subscripts are mole ratios. Four steps do it every time.

Start with grams. If you were given percents, pretend you hold exactly 100.00 g of the compound. Then each percent is already a mass in grams.
Divide each mass by that element's molar mass. Now every number is in moles.
Divide every mole value by the smallest of them. The smallest becomes 1.000 and the rest fall close to whole numbers.
If a value sits near .50, .33 or .25, multiply all of them by 2, 3 or 4 to clear it.

Judgment lives in step 4. A value of 1.99 is 2, since lab data is never perfect. A value of 1.50 is not 2. Rounding it would change the compound. Multiply instead.

Scaling up to the molecule

The empirical formula has a mass of its own, called the empirical formula mass. Divide the real molar mass by it. The answer is a whole number n, the number of empirical units per molecule. Multiply every subscript by n and you have the molecular formula.

n = molar mass ÷ empirical formula mass

If n comes out as 1, the two formulas are the same. Water and ammonia are both like that. If n lands on 2.98 or 3.03, treat it as 3 and move on.

Figure 3.4 The route from analysis to a formula. Percent data stops at the empirical formula. Only a measured molar mass gets you the last box.

Worked example · percents to a molecular formula

A compound is 40.00% C, 6.71% H and 53.28% O by mass. A separate measurement puts its molar mass at 180.16 g/mol. Find both formulas.

Step 1 · assume 100.00 gThat gives 40.00 g C, 6.71 g H and 53.28 g O.

Step 2 · grams to molesC: 40.00 / 12.011 = 3.330 mol. H: 6.71 / 1.008 = 6.657 mol. O: 53.28 / 15.999 = 3.330 mol.

Step 3 · divide by the smallestThe smallest is 3.330. C: 3.330 / 3.330 = 1.000. H: 6.657 / 3.330 = 1.999. O: 3.330 / 3.330 = 1.000.

Step 4 · read the empirical formulaThe ratio is 1 : 2 : 1, so the empirical formula is CH₂O.

Step 5 · empirical formula mass12.011 + 2(1.008) + 15.999 = 30.026, or 30.03 g/mol.

Step 6 · find nn = 180.16 / 30.03 = 5.999, which is 6.

Step 7 · scale upMultiply each subscript by 6. The molecular formula is C₆H₁₂O₆.

Worked example · masses to a formula

A sample of an iron oxide is broken down and found to hold 11.17 g of Fe and 4.80 g of O. What is its formula?

Step 1 · grams to molesFe: 11.17 / 55.845 = 0.2000 mol. O: 4.80 / 15.999 = 0.3000 mol.

Step 2 · divide by the smallestFe: 0.2000 / 0.2000 = 1.000. O: 0.3000 / 0.2000 = 1.500.

Step 3 · clear the half1.500 is not close to a whole number. Multiply both by 2. Fe becomes 2 and O becomes 3.

Step 4 · write itThe formula is Fe₂O₃, iron(III) oxide, at 159.69 g/mol. It is ionic, so this is both its empirical and its only formula.

Key points

An empirical formula is the smallest whole-number atom ratio. A molecular formula is the real count per molecule.
Subscripts are mole ratios, so always convert mass to moles first.
Percents alone give the empirical formula. Molar mass is needed for the molecular one.
Divide by the smallest mole value, then multiply out any stubborn halves or thirds.
n = molar mass divided by empirical formula mass, and n is always a whole number.

Interactive Turn percent data into an empirical formula, then a molecular one. Open full ↑

Section 5Molarity

Strength is not amount

A solution is a mixture so evenly blended that it acts like one substance. The part that dissolves is the solute. The part it dissolves into is the solvent. When the solvent is water we call it an aqueous solution and tag the formula with (aq).

Picture two glasses of the same salt water. One holds a sip and one holds a liter. The amount of salt in them is very different. The strength is identical. Concentration is the word for that strength. It is the amount of solute per amount of solution, and it does not care how much you poured.

Molarity is the concentration unit chemistry leans on hardest. It is moles of solute per liter of solution. The symbol is a capital M. A bottle marked 2.0 M is read out loud as two molar.

molarity (M) = moles of solute ÷ liters of solution

Of solution, not of solvent

Read the bottom of that fraction again. It says liters of solution. It does not say liters of solvent. The difference is real work in the lab. You do not pour 1.00 L of water onto your solid. You drop the solid into a volumetric flask, swirl it into part of the water, then top up to the etched mark on the neck. The solute takes up room of its own. Filling to the mark is the only way to know your final volume.

Molarity has one soft spot. Liquids swell when warmed, so a volume shifts a little with temperature. A 0.100 M solution on a hot bench is a hair weaker than the same flask in a cold room. For ordinary lab work the drift is too small to matter.

Three ways the equation gets used

Molarity, moles and volume form a small triangle. Know any two and you can get the third. Moles = M × V. Volume = moles ÷ M. Convert the volume to liters before anything else. 250.0 mL is 0.2500 L, and 25.0 mL is 0.0250 L.

Figure 3.5 Molarity counts liters of finished solution, not liters of water. The flask is filled to the etched mark after the solid has dissolved.

Worked example · molarity from a weighed mass

You dissolve 11.7 g of NaCl and top the flask up to 250.0 mL. What is the molarity?

Step 1 · grams to molesNaCl is 58.44 g/mol. 11.7 g × (1 mol / 58.44 g) = 0.2002 mol.

Step 2 · volume to liters250.0 mL × (1 L / 1000 mL) = 0.2500 L.

Step 3 · divide0.2002 mol ÷ 0.2500 L = 0.8008 mol/L.

Step 4 · roundThe mass had three figures, so the answer is 0.801 M.

Worked example · working back to a mass

You need 500.0 mL of 0.150 M glucose. How much glucose do you weigh out?

Step 1 · find the molesmoles = M × V = 0.150 mol/L × 0.5000 L = 0.0750 mol.

Step 2 · moles to grams0.0750 mol × 180.16 g/mol = 13.512 g, so 13.5 g.

Step 3 · make it upWeigh 13.5 g into a 500.0 mL volumetric flask. Swirl it into some water. Top up to the mark and invert to mix.

Step 4 · one more useHow many moles sit in a 25.0 mL pipette of 0.100 M HCl? 0.0250 L × 0.100 mol/L = 2.50 × 10^-3 mol.

Key points

Molarity is moles of solute per liter of solution, written M.
The volume is the finished solution, measured to the mark, not the water you started with.
moles = M × V, and V = moles ÷ M. Put volume in liters first.
To build a solution from a solid, find the moles you need, then convert to grams.
Equal volumes of the same molarity hold equal moles of solute.

Interactive Make a solution to a target molarity and check your work. Open full ↑

Section 6Dilution

Add water, keep the solute

Labs rarely weigh out a solid for every weak solution they need. They mix a strong one and water it down. The strong one is the stock solution. Watering it down is dilution.

One idea makes every dilution problem easy. Adding water adds no solute and removes none. The moles before equal the moles after. Only the volume grew, so the concentration fell.

Moles are M × V, so that sentence becomes an equation. Subscript 1 is before and 2 is after.

M₁V₁ = M₂V₂

Both sides of that line are moles of solute. The same volume unit appears on each side, so you may leave both volumes in milliliters. The only rule is to use one unit throughout.

Working out the pour

Usually you know the stock concentration, the one you want and the final volume. The unknown is how much stock to measure out. Rearrange for V₁.

V₁ = M₂V₂ ÷ M₁

Measure out that much stock, then add solvent up to V₂. With a strong acid the order matters. Acid goes into water, never water into acid, because mixing releases a lot of heat.

The dilution factor

The dilution factor is V₂ ÷ V₁. It also equals M₁ ÷ M₂, a handy second check. Take 10.0 mL up to 100.0 mL and the factor is 10. The concentration drops to one tenth. That is a 1:10 dilution.

Factors multiply when you chain steps. Two 1:10 steps in a row give a factor of 100. This is called a serial dilution. It is how a lab reaches very low concentrations without weighing out a speck of solid.

Figure 3.6 Both beakers hold the same eight particles of solute. Only the volume changed, so the drop in molarity matches the growth in volume exactly.

Worked example · how much stock to pour

You need 500.0 mL of 0.500 M HCl. The shelf holds 12.0 M HCl. How much stock do you measure?

Step 1 · list what you knowM₁ = 12.0 M. M₂ = 0.500 M. V₂ = 500.0 mL.

Step 2 · rearrangeV₁ = M₂V₂ ÷ M₁.

Step 3 · put the numbers inV₁ = (0.500 M × 500.0 mL) ÷ 12.0 M = 250 ÷ 12.0 = 20.83 mL, so 20.8 mL.

Step 4 · check both sidesStock side: 12.0 mol/L × 0.0208 L = 0.250 mol. Final side: 0.500 mol/L × 0.5000 L = 0.250 mol. The two agree.

Step 5 · do it safelyAdd the acid to water already in the flask, then top up to 500.0 mL.

Worked example · the new concentration

You pipette 25.0 mL of 0.800 M KCl into a flask and top up to 200.0 mL. Find the new molarity.

Step 1 · rearrange for M₂M₂ = M₁V₁ ÷ V₂.

Step 2 · substituteM₂ = (0.800 M × 25.0 mL) ÷ 200.0 mL = 20.0 ÷ 200.0 = 0.100 M.

Step 3 · check with the factorThe volume grew 200.0 / 25.0 = 8.00 times, so the molarity should fall 8.00 times. 0.800 ÷ 8.00 = 0.100 M. It agrees.

Key points

Dilution adds solvent only. The moles of solute never change.
M₁V₁ = M₂V₂, with before on the left and after on the right.
Both volumes may stay in milliliters, if you use one unit on both sides.
V₁ = M₂V₂ ÷ M₁ tells you how much stock to measure.
The dilution factor is V₂/V₁ and also M₁/M₂. Chained steps multiply.
Pour strong acid into water, not water into acid.

Interactive Dilute a stock solution with M1V1 = M2V2. Open full ↑

The chapter in one breath

A mole is 6.022 × 10²³ particles. It is a counting word, like a dozen, sized for atoms.
Molar mass in g/mol is the sum of the atomic masses in a formula. It links grams to moles.
Avogadro's number links moles to a count. Every problem runs through moles in the middle.
Percent composition is each element's mass in one mole divided by the molar mass, times 100.
Percent composition belongs to the compound, so sample size never enters the math.
An empirical formula is the smallest whole-number atom ratio. Convert mass to moles, then divide by the smallest.
Molar mass divided by empirical formula mass gives n, and n scales the ratio up to the molecular formula.
Molarity is moles of solute per liter of solution, measured to the mark in a volumetric flask.
Dilution keeps the solute and grows the volume, so M₁V₁ = M₂V₂.

ReferenceWords worth knowing

These are the terms this chapter leans on. Each one shows up in the worked examples above, so read them with those numbers in mind.

Aqueous solution: A solution whose solvent is water. Written with the tag (aq) after a formula.
Avogadro's number: The number of particles in one mole, N_A = 6.022 × 10²³ per mole. Fixed by definition at 6.02214076 × 10²³.
Concentration: The amount of solute per amount of solution. A measure of strength, not of quantity.
Dilution: Adding solvent to a solution to lower its concentration. The moles of solute stay the same.
Dilution factor: V₂/V₁, equal to M₁/M₂. A factor of 10 means a 1:10 dilution.
Empirical formula: The smallest whole-number ratio of atoms in a compound. CH₂O is the empirical formula of glucose.
Formula mass: The sum of the atomic masses in a formula, in amu. Numerically equal to the molar mass in g/mol.
Formula unit: The smallest neutral group an ionic compound's formula describes, such as one Na⁺ with one Cl^-.
Hydrate: A solid that holds water inside its crystal in a fixed ratio, shown with a raised dot, as in CuSO₄·5H₂O.
Mass percent: The share of a compound's mass owned by one element, as a percent. Also called percent by mass.
Molar mass: The mass of one mole of a substance, in grams per mole. The bridge between grams and moles.
Molarity: Moles of solute per liter of solution, symbol M. The most used concentration unit in this book.
Mole: The counting unit for particles. One mole is 6.022 × 10²³ atoms, molecules or formula units.
Molecular formula: The actual number of atoms of each element in one molecule, such as C₆H₁₂O₆.
Percent composition: The full set of mass percents for a compound. The shares add to 100%.
Solute: The substance that dissolves in a solution. It is usually the smaller part by mass.
Solution: A mixture blended so evenly that it behaves as a single substance.
Solvent: The substance that does the dissolving. Water is the most common one.
Stock solution: A concentrated solution kept on hand and diluted when a weaker one is needed.
Volumetric flask: A flask with one etched line on its neck, used to make a solution up to an exact volume.

Check yourself

1. How many moles are in 54.0 g of water?

3.00 mol Water is 2(1.008) + 15.999 = 18.015, so 18.02 g/mol. Then 54.0 g × (1 mol / 18.02 g) = 2.997 mol, which rounds to 3.00 mol. That is also 3.00 × 6.022 × 10²³ = 1.81 × 10²⁴ molecules.

2. A friend says a mole of anything weighs the same. What is wrong with that?

A mole is a count, not a mass. One mole always holds 6.022 × 10²³ particles, but those particles have different masses. A mole of water weighs 18.02 g while a mole of glucose weighs 180.16 g.

3. Find the molar mass of Al₂(SO₄)₃.

342.13 g/mol The 3 outside multiplies the whole sulfate group, so there are 2 Al, 3 S and 12 O. Al: 2 × 26.982 = 53.964. S: 3 × 32.06 = 96.18. O: 12 × 15.999 = 191.988. Sum = 342.132 g/mol.

4. What percent of the mass of CO₂ is oxygen?

72.71% CO₂ is 12.011 + 2(15.999) = 44.009, so 44.01 g/mol. Oxygen contributes 31.998 g per mole. (31.998 / 44.009) × 100 = 72.71%. Carbon takes the other 27.29%.

5. A manganese oxide is 69.6% Mn and 30.4% O. What is its empirical formula?

Mn₂O₃ Assume 100.00 g. Mn: 69.6 / 54.938 = 1.267 mol. O: 30.4 / 15.999 = 1.900 mol. Divide both by 1.267. Mn = 1.000 and O = 1.500. Multiply by 2 to clear the half, giving Mn₂O₃.

6. How many grams of KMnO₄ make 100.0 mL of 0.0200 M solution?

0.316 g moles = 0.0200 mol/L × 0.1000 L = 2.00 × 10^-3 mol. KMnO₄ is 39.098 + 54.938 + 4(15.999) = 158.032, so 158.03 g/mol. Mass = 2.00 × 10^-3 mol × 158.03 g/mol = 0.316 g.

7. What volume of 6.00 M NaOH gives 250.0 mL of 0.150 M NaOH?

6.25 mL Use V₁ = M₂V₂ ÷ M₁ = (0.150 × 250.0) ÷ 6.00 = 37.5 ÷ 6.00 = 6.25 mL. Measure that out, then add water up to 250.0 mL.

8. Why does molarity use liters of solution rather than liters of solvent?

Because the solute takes up room too. If you added one liter of water to a solid, the finished volume would be more than one liter. Topping up to a flask's mark is the only way to fix the volume you divide by.

9. Benzene is C₆H₆. What is its empirical formula, and why is that not enough to identify it?

CH, and because many compounds share it. Dividing both subscripts by 6 gives CH. Acetylene, C₂H₂, reduces to CH as well. Only the molar mass tells the two apart, since n = 78.11 / 13.02 = 6 for benzene and 2 for acetylene.

10. How many oxygen atoms are in 1.50 mol of Ca(NO₃)₂?

5.42 × 10²⁴ atoms Each formula unit holds 6 O, so 1.50 mol × 6 = 9.00 mol O. Then 9.00 mol × 6.022 × 10²³/mol = 5.42 × 10²⁴ oxygen atoms.

11. You dilute 50.0 mL of 0.200 M KCl to 500.0 mL. How many moles of KCl are in the flask afterward?

0.0100 mol, the same as before. Dilution adds no solute. Before: 0.0500 L × 0.200 mol/L = 0.0100 mol. After: M₂ = (0.200 × 50.0) / 500.0 = 0.0200 M, and 0.5000 L × 0.0200 mol/L = 0.0100 mol.

12. Formaldehyde, acetic acid and glucose all give 40.00% C, 6.71% H and 53.28% O. What extra fact separates them?

The molar mass. All three share the empirical formula CH₂O, whose mass is 30.03 g/mol. Formaldehyde is 30.03 g/mol so n = 1. Acetic acid is 60.05 g/mol so n = 2. Glucose is 180.16 g/mol so n = 6.

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