Percentage composition · printable packet

What share of a compound's mass each element accounts for, and why that is not the share of its atoms.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to work out the percentage by mass of any element in a compound — its contribution to one mole divided by the molar mass — and to use that percentage the other way, to find the mass of an element in a weighed sample. You will be able to say why the percentage does not depend on how much you have, and why a share of the mass is a different number from a share of the atoms.

2. What you already have

You can work out a molar mass, and in doing so you already work out what each element contributes to it — that is the middle step of the calculation. A percentage composition takes those contributions and expresses each as a share of the whole.

3. Words for this lesson

4. A share of the mass

$$\text{percentage by mass} = \frac{\text{mass of that element in one mole}}{\text{molar mass}} \times 100.$$

Both numbers come out of the molar mass calculation you already do. For $\mathrm{CaCO_3}$, molar mass 100: calcium contributes 40, carbon 12, oxygen 48. So the composition is 40.0% calcium, 12.0% carbon, 48.0% oxygen, and those add to 100 as they must.

Term	What it means
Percentage composition by mass	For each element, the share of the compound's mass it accounts for, as a percentage.
Contribution	An element's atomic mass times its number of atoms in the formula.
Tolerance	How far an answer may be from the stated value and still be accepted.
Significant figures	The digits of a number that carry measured information.
Ore	A rock containing enough of a metal compound to be worth mining.

It does not depend on how much you have. A percentage is a ratio, and the amount cancels. A ton of calcium carbonate is 40% calcium; so is a milligram. That is what makes it a property of the substance rather than of the sample.

It is a share of the mass, not of the atoms. In ammonia, $\mathrm{NH_3}$, one atom in four is nitrogen — 25% of the atoms. But nitrogen weighs 14 and hydrogen weighs 1, so nitrogen is $14/17 = 82.4\%$ of the mass. Both numbers are true; only one of them is the percentage composition.

Another way: picture

Picture one mole as a bar of length equal to the molar mass, cut into pieces — one piece per element, each as long as that element's contribution. The percentage composition is the length of each piece as a share of the whole bar.

Another way: steps

To find the percentage of an element:

Work out the molar mass of the compound.
Take the part of it that came from that element: its atomic mass times how many atoms of it the formula has.
Divide, multiply by 100, and quote to one decimal place.

5. The method, step by step, and how to check it

Read the formula. Count the atoms of the element you want, including any inside brackets or attached to a water of crystallization.

Work out its contribution. Atomic mass times that count. This is the number that most often goes wrong, because it is tempting to use the atomic mass alone.

Divide and scale. Contribution over molar mass, times 100. Quote to one decimal place.

If a sample mass is given, multiply it by the percentage as a decimal — or convert the sample to moles and multiply by the contribution. Both routes give the same answer.

Check the work. Three tests catch nearly every error. Do the percentages of all the elements add to 100, within a tenth or two of rounding? Is a heavy element's share larger than its share of the atoms, and a light element's — hydrogen especially — smaller? And is every percentage between 0 and 100? An answer above 100 means the division was upside down.

6. Why each step is allowed

Using one mole as the sample is allowed because the percentage does not depend on the amount. Any sample would give the same ratio; one mole is chosen because its masses are the numbers already worked out for the molar mass.

Multiplying the atomic mass by the subscript is required because the element's share of the mass is the mass of all its atoms, not of one. Two iron atoms in hematite weigh twice what one does.

Dividing by the whole molar mass is right because a percentage is always a share of a whole, and the whole here is the mass of the entire formula unit.

Taking a percentage of a sample's mass to find an element's mass is allowed because the ratio is the same in every sample of a pure compound — the law of definite proportions, which says a compound always contains its elements in the same proportions by mass, however it was made.

7. Why every sample gives the same answer

Around 1800 the French chemist Joseph Proust analyzed copper carbonate made in his laboratory and copper carbonate dug from the ground, and found the same proportions of copper, carbon and oxygen by mass in both, to the limits of his balance. He did the same with other compounds and reached the same result every time. The rule became known as the law of definite proportions: a pure compound contains its elements in fixed proportions by mass, wherever it came from and however it was made.

Today that looks obvious, because a formula says it directly. Every unit of $\mathrm{CaCO_3}$ has one calcium, one carbon and three oxygens, so every sample has the same shares by mass. But the law came first, and it was one of the strongest pieces of evidence that matter is made of atoms combining in whole-number ratios.

It is also why a percentage composition is useful in practice. A quality lab can test a few grams of a shipment and know the composition of every ton of it, as long as the material is pure. When the measured percentage differs from the formula's, the difference itself is information: it says the sample contains something else, and how much.

8. Going the other way: how much of an element is in a sample

Because the percentage does not depend on the amount, it works as a conversion factor in both directions.

A 250 g sample of calcium carbonate contains $40\%$ of 250 g of calcium, which is 100 g.

The longer route gives the same answer and is worth doing once: 250 g is 2.5 mol, each mole carries 40 g of calcium, so $2.5 \times 40 = 100$ g. The two routes are the same arithmetic with the division moved, and which one is quicker depends on whether the numbers happen to divide nicely.

This is exactly the calculation a mining company does to decide whether an ore is worth processing, and a farmer does to compare two sacks of fertilizer.

9. The rounding rule, once

Percentages almost never come out exact. $12/44$ is $27.2727\ldots$, and there is no honest way to write it in full.

So this course states every percentage to one decimal place, and every item that asks for one accepts anything within $0.1$ of the stated value. An answer of $27.27$ and an answer of $27.3$ are both correct, and so is $27.2$.

A tolerance and a precision rule are two different things, and this lesson uses one of each so that the difference is visible. A tolerance says near enough: write as many digits as you like, and anything close enough passes. A precision rule says this many digits, and marks the digits themselves.

The percentage questions here use a tolerance. The question that asks for the mass of an element in a weighed sample uses a precision rule — three significant figures, the same three the sample was weighed to — and it says so in the question, because being marked on digits nobody asked for would be unfair. Quoting an answer more precisely than the measurement it came from claims accuracy that was never there; quoting it less precisely throws away a digit that was.

10. In the world: the three numbers on a fertilizer bag

Every bag of fertilizer sold in the United States carries three numbers by law, such as 46-0-0 or 10-10-10. They are percentages by mass: nitrogen, then phosphate, then potash. A bag of pure urea, $\mathrm{CO(NH_2)_2}$, reads 46-0-0, and this lesson's arithmetic is where that 46 comes from: two nitrogen atoms contribute $2 \times 14 = 28$ g to a molar mass of 60 g/mol, and $28 \div 60 \times 100 = 46.7\%$, which the label rounds down.

A farmer in Iowa planning corn needs roughly 150 pounds of nitrogen per acre. The percentage turns that into a purchase. Urea at 46% needs $150 \div 0.46 \approx 326$ pounds of product per acre. Ammonium nitrate, $\mathrm{NH_4NO_3}$, is 35% nitrogen, so it needs $150 \div 0.35 \approx 429$ pounds — a third more to haul and spread for the same nitrogen.

That is why urea is the most widely used solid nitrogen fertilizer in the country: it carries more of the element that matters per ton shipped. The comparison rests entirely on percentage by mass. By atoms, urea's nitrogen is 2 of 8, just 25%, a number that says nothing about how many pounds of nitrogen reach the field. State agriculture departments test bags to confirm the printed figure, because a farmer paying for 46% and receiving 40% is paying for nitrogen they never got.

11. In the world: the nutrition label on a salt shaker

Table salt is 39.3% sodium by mass: $23 \div 58.5 \times 100$. That is why the 2300 mg daily sodium limit in federal dietary guidelines works out to about 5.8 g of salt, a little over one teaspoon.

12. Where this goes wrong

Counting atoms instead of mass. The error this lesson exists for. In water, hydrogen is two atoms out of three — and $11.1\%$ of the mass, because a hydrogen atom weighs a sixteenth of what an oxygen atom does. If the answer you have is a simple fraction of the atom count, check which question you answered.

Using the atomic mass instead of the element's contribution. For $\mathrm{H_2SO_4}$, hydrogen contributes 2 to the molar mass, not 1. Dropping the subscript here is the same error as dropping it in the molar mass, arriving one step later.

Dividing by the wrong thing. The divisor is always the whole molar mass. Dividing by another element's contribution gives a ratio between two elements, which is a real and different quantity.

Forgetting the water of crystallization. In $\mathrm{CuSO_4 \cdot 5H_2O}$ the percentage of copper is worked out against $249.5$, not $159.5$. The water is part of what you weighed.

Rounding too early. Rounding the molar mass or a contribution before dividing can move the last digit of the percentage. Keep every figure until the final division, then round once.

13. Iron in its ore

14. Where by-mass and by-atoms disagree most

15. Copper in a weighed sample of crystals

16. Your turn: what percentage of magnesium oxide, $\mathrm{MgO}$, is oxygen by mass?

17. Guided practice

Three numbers have been worked out for methane, $\mathrm{CH_4}$. Which one is the percentage of carbon **by mass**?

18. Guided practice

Complete the worked solution: hematite is $\mathrm{Fe_2O_3}$. Use iron at fifty-six and oxygen at sixteen. Find iron's contribution to one mole, the molar mass, and the percentage of iron by mass.

19. Guided practice

What percentage of the mass of ammonia, $\mathrm{NH_3}$, is nitrogen? Give your answer to one decimal place.

20. Practice

A sample of glucose, $\mathrm{C_6H_{12}O_6}$, is weighed and comes to $360$ g. What mass of carbon does it contain? Answer in grams, to three significant figures.

21. Practice

A steel company buying ore from Minnesota's Iron Range prices it on its iron content. Pure hematite is $\mathrm{Fe_2O_3}$. Using iron at 56, carbon at 12 and oxygen at 16, what percentage of it is iron by mass? Give one decimal place.

22. Somewhere new

Fertilizer is sold on one number: the percentage of nitrogen it contains by mass, printed on the sack. A sack of potassium nitrate, $\mathrm{KNO_3}$, is being labeled. Work out the three figures behind the number on the label.

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

The compound is magnesium hydroxide, $\mathrm{Mg(OH)_2}$, and its molar mass is $58$ g/mol. The mass each element contributes to one mole is given. Work out what percentage of the compound's mass each element is, to one decimal place.

25. What you can do now

You can find the percentage of an element by mass, and the mass of it in a sample. Say what percentage of methane is carbon by mass, and why it is not the twenty percent you get by counting atoms. Next: running the calculation backwards — from a set of percentages to the formula that produced them.

Working for the steps left to you

	mass of nitrogen in one mole, in g	molar mass, in g per mol	nitrogen by mass, as a percentage
$\mathrm{KNO_3}$

	mass in one mole, in g	percentage by mass
$\mathrm{Mg}$	24
$\mathrm{O}$	32
$\mathrm{H}$	2

16. Your turn: what percentage of magnesium oxide, $\mathrm{MgO}$, is oxygen by mass?, step 3

$16 \div 40 \times 100 = 40.0\%$

Magnesium is the other 60.0%.