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Turning percentages by mass into a ratio of atoms, and what a molar mass adds that composition never can.
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.
By the end of this lesson you will be able to work out an empirical formula from a percentage composition — treating the percentages as masses in 100 g, dividing each by the element's relative atomic mass, then reducing to the simplest whole-number ratio and doubling rather than rounding a half. You will also be able to turn an empirical formula into a molecular one when a molar mass is supplied, and to say why composition alone can never do that.
You can turn a formula into a percentage composition. This lesson goes the other way, and the only new idea is that a percentage has to be converted from a share of mass into a count before it can become a subscript.
| Term | What it means |
|---|---|
| Empirical formula | The simplest whole-number ratio of the atoms in a compound. |
| Molecular formula | How many atoms of each element are actually in one molecule. |
| Empirical formula mass | The mass of one empirical unit, worked out like a molar mass. |
| Combustion analysis | Burning a sample and weighing the carbon dioxide and water it makes. |
| Mass spectrum | A chart from an instrument that sorts ions by mass, showing the molecule's mass. |
Every formula in this course so far arrived already written. Somebody had to find the first one, and all they had was a balance.
A sample of an unknown compound is burned and the products are trapped and weighed. The analysis reports only this:
| Mass in a 100 g sample | |
|---|---|
| carbon | 75.0 g |
| hydrogen | 25.0 g |
The question is: what is the formula?
Here is the trap, and it catches nearly everybody, so commit to an answer before you read further. There is three times as much carbon as hydrogen by mass. Does that make the formula $\mathrm{C_3H}$?
Test it against something you can already check. Work out the percentage composition of $\mathrm{C_3H}$ the way you did last lesson, and see whether it comes to 75 and 25. It does not. So the shares of mass are not the subscripts, and the reason is in one line of the table you already have: a carbon atom is twelve times as heavy as a hydrogen atom. 75 grams of carbon and 25 grams of hydrogen are not 75 atoms and 25 atoms.
So: what would you have to do to each mass to turn it into a number of atoms? You know what one atom of each weighs, relative to the others. Say the operation, do it to both numbers, and see whether what comes out looks like a formula.
One more thing to decide while you are there. Suppose two different substances — a welding gas and a liquid solvent — both analyze as 92.3% carbon and 7.7% hydrogen. Can any amount of weighing the burned products tell them apart? If not, what other measurement would?
A percentage composition is a share of the mass. A formula is a ratio of counts. The whole method is the conversion between those two, and it is one division.
Dividing a mass by the mass of one atom of that element tells you how many atoms there were. So:
That gives the empirical formula — and it is all that composition can ever give, because $\mathrm{CH_2}$ and $\mathrm{C_2H_4}$ and $\mathrm{C_3H_6}$ have identical percentage compositions.
To get the molecular formula you need one more fact: the molar mass, usually from a mass spectrum. Divide it by the empirical unit's mass, and multiply every subscript by the whole number that comes out.
Another way: picture
Picture two piles of hardware, weighed rather than counted: a pile of bolts and a pile of nuts. Knowing the weight of each pile tells you nothing about how many objects are in it until you know what one object weighs. Dividing by the atomic mass is weighing one object.
Another way: steps
75.0% carbon, 25.0% hydrogen:
Assume 100 g. Each percentage becomes a mass in grams. If the question gives masses instead of percentages — grams of each element from an actual sample — use those directly; the method does not care which.
Divide by atomic masses. Each mass over the atomic mass of its element, kept to two decimal places. These are now relative counts of atoms.
Divide by the smallest. Every count over the smallest one, so that one of them becomes exactly 1.
Clear fractions. Close to a whole number: round. Close to a half: double everything. Close to a third or two thirds: triple everything. Close to a quarter: multiply by four.
Write the empirical formula, then, if a molar mass is given, divide it by the empirical formula mass and multiply every subscript by the result.
Check the work. Work out the percentage composition of your answer, as in the last lesson, and compare it with the numbers you started from; they should agree to within the rounding. Check that the multiple from the molar mass is a whole number. And check that the formula makes chemical sense: a compound of carbon and hydrogen with more carbon atoms than hydrogen atoms is possible, but one with ten times as many is a sign of a dropped division.
Assuming 100 g is allowed because the formula does not depend on the sample size. Any mass would give the same ratio; 100 g is chosen because it makes the percentages into grams with no arithmetic.
Dividing by the atomic mass is allowed because the atomic mass is the relative mass of one atom. A total mass divided by the mass of one item is the number of items, whether the items are bolts or carbon atoms.
Dividing every count by the smallest is allowed because a ratio is unchanged when every part is divided by the same number, and it is what makes the answer recognizable.
Multiplying to clear a half is allowed for the same reason, in the other direction: scaling every part of a ratio leaves the ratio the same, and atoms come only in whole numbers.
Using the molar mass to find the multiple is allowed because the molecule is a whole number of empirical units, so its mass is that whole number times the unit's mass.
Dividing both counts by the smaller of them looks like a trick and is not. A formula is a ratio, and a ratio is unchanged by dividing both sides by the same number — so the division is free, and choosing the smallest number as the divisor is what makes one of the results exactly 1.
That matters because it is the only way to recognize the answer. $6.25$ and $25$ could be almost any ratio; $1$ and $4$ is unmistakably $\mathrm{CH_4}$.
When the result is a half, double both. $1 : 1.5$ is $2 : 3$, which is $\mathrm{Fe_2O_3}$ and not a mistake. A third means multiply by three. Anything within a few hundredths of a whole number is that whole number, because the percentages you started from were rounded and the small discrepancy came from the rounding rather than from the chemistry.
| Empirical formula | Formula mass | Molar mass | Molecular formula | |
|---|---|---|---|---|
| ethyne | $\mathrm{CH}$ | 13 | 26 | $\mathrm{C_2H_2}$ |
| benzene | $\mathrm{CH}$ | 13 | 78 | $\mathrm{C_6H_6}$ |
| hydrogen peroxide | $\mathrm{HO}$ | 17 | 34 | $\mathrm{H_2O_2}$ |
The first two rows are the point of the table. Ethyne is a welding gas and benzene is a liquid solvent; they have the same empirical formula and identical percentage compositions, and no amount of analyzing their composition will ever tell them apart.
The molar mass is what separates them, and it comes from a different instrument entirely. That is a general shape worth noticing: two kinds of evidence, each blind to what the other sees, and the identification needs both.
When police in an American city seize unlabeled pills, a state crime laboratory has to say what is in them before a case can go to court, and the method is this lesson run with modern instruments. An elemental analyzer burns a few milligrams of the powder in pure oxygen and measures the carbon dioxide, water and nitrogen that come off. From those it reports the percentages of carbon, hydrogen and nitrogen, and oxygen by difference.
The analyst turns those percentages into an empirical formula exactly as in this lesson: divide by atomic masses, divide by the smallest, clear any fractions. Suppose the result is $\mathrm{C_4H_5N_2O}$, with a unit mass of 97. On its own, that formula fits caffeine, but it would also fit any molecule twice or three times as large.
So the same sample goes through a mass spectrometer, which reports a molecular mass of 194. $194 \div 97 = 2$, giving $\mathrm{C_8H_{10}N_4O_2}$: caffeine, not a controlled substance. Had the spectrum shown a different mass, or the percentages a different ratio, the identification would have changed with it. Courts accept this kind of evidence because the two measurements are independent — composition from one instrument, size from another — and each checks a different thing. Forensic chemists testify to exactly the reasoning in this lesson, and a defense attorney who knows it can ask the questions that test whether it was done right.
The FDA tests supplements for what their labels claim. A vitamin C tablet whose analysis gave anything other than $\mathrm{C_3H_4O_3}$ as the empirical formula would signal a filler or a substitute, before any further test was run.
Using the percentages as subscripts. 75% carbon and 25% hydrogen read as $\mathrm{C_{75}H_{25}}$, or its simplification $\mathrm{C_3H}$. Both are backwards: the percentages are masses, and a formula counts atoms. The division by the atomic mass is the entire method and it cannot be skipped.
Dividing by the largest instead of the smallest. That gives a ratio with a number below 1 in it, and nothing whole to recognize. Always divide by the smallest.
Rounding a genuine half. $1 : 1.5$ rounded to $1 : 2$ turns iron(III) oxide into iron(II) oxide. A half is doubled, not rounded. A value of $1.98$, on the other hand, is 2 — the percentages had already been rounded, and that is where the two hundredths came from.
Stopping at the empirical formula. If the question gives you a molar mass, it is giving it to you for a reason. Check whether the empirical unit's mass already equals it, and if it does not, the multiple is what you were asked for.
Rounding the counts too early. Rounding 1.875 to 2 before dividing turns a clear $1 : 1.5$ into $1 : 1.6$, which looks like nothing. Keep two decimal places until the last step.
Take a 100 g sample.
$70 \text{ g Fe}, \ 30 \text{ g O}$
Percentages become masses.
Count the iron atoms.
$70 \div 56 = 1.25$
Mass over one atom's mass.
Count the oxygen atoms.
$30 \div 16 = 1.88$
The same conversion.
Divide by the smaller.
$1 : 1.5$
A half has appeared.
Double to clear it.
$2 : 3 \Rightarrow \mathrm{Fe_2O_3}$
Rounding would give FeO, a real and wrong compound.
Count the carbon atoms.
$85.7 \div 12 = 7.14$
From 85.7 g carbon.
Count the hydrogen atoms.
$14.3 \div 1 = 14.3$
From 14.3 g hydrogen.
Divide by the smaller.
$1 : 2.00 \Rightarrow \mathrm{CH_2}$
The empirical formula.
Weigh one empirical unit.
$12 + 2 = 14$
Composition can say no more.
Divide the molar mass by it.
$28 \div 14 = 2$
The spectrum gives 28.
Multiply every subscript.
$\mathrm{C_2H_4}$
Ethene, the gas used to ripen fruit.
Take a 100 g sample.
$40.9 \text{ g C}, \ 4.58 \text{ g H}, \ 54.5 \text{ g O}$
Three elements, same method.
Count the carbon atoms.
$40.9 \div 12 = 3.41$
Mass over atomic mass.
Count the hydrogen atoms.
$4.58 \div 1 = 4.58$
Hydrogen's mass is one.
Count the oxygen atoms.
$54.5 \div 16 = 3.41$
The same as carbon.
Divide by the smallest.
$1 : 1.34 : 1$
A third has appeared.
Triple to clear it.
$3 : 4 : 3 \Rightarrow \mathrm{C_3H_4O_3}$
The empirical formula, mass 88.
Use the molar mass.
$176 \div 88 = 2 \Rightarrow \mathrm{C_6H_8O_6}$
Ascorbic acid's molecular formula.
Count each kind of atom.
$60 \div 24 = 2.5; \ 40 \div 16 = 2.5$
Masses to counts.
Divide by the smaller.
$1 : 1$
The two counts are equal.
Write the formula.
Analysis of a gas that ripens fruit gives the empirical formula $\mathrm{CH_2}$, and a mass spectrum gives a molar mass of $28$ g/mol. Which is the molecular formula?
Complete the worked solution: a rust sample is seventy percent iron and thirty percent oxygen by mass. Use iron at fifty-six and oxygen at sixteen. Find the iron count, the oxygen count, and the oxygen-to-iron ratio.
Count the iron atoms.
$\text{seventy} \div \text{fifty-six} =$ f
Percent as grams in a hundred-gram sample.
Count the oxygen atoms.
$\text{thirty} \div \text{sixteen} =$ o
The same conversion.
Divide by the smaller count.
$(\text{oxygen}) \div (\text{iron}) =$ r
A half means double: two to three.
The sample is a solvent smelling of gasoline. It has the empirical formula $\mathrm{CH}$, whose unit mass is $13$, and a molar mass of $78$ g/mol. How many empirical units are there in one molecule?
Answer:
The atoms in the brown gas above a busy road are in the ratio $1$ $\mathrm{N}$ to $2$ $\mathrm{O}$. Write the empirical formula, and the mass of one empirical unit. Type a formula with no subscript formatting, as in CH4 or Fe2O3.
The empirical formula of the brown gas above a busy road is a, and one empirical unit has a mass of b.
An FDA laboratory checks a sample of caffeine from an energy drink. Combustion analysis gives the empirical formula $\mathrm{C_4H_5N_2O}$, unit mass $97$, and the mass spectrometer gives a molar mass of $194$ g/mol. How many carbon atoms are in one molecule?
The answer: a.
A forensic laboratory has an unidentified sample. Combustion analysis gives it the empirical formula $\mathrm{HO}$, and the heaviest peak in its mass spectrum sits at $34$ — which for this sample is the mass of the whole molecule. Identify the compound. Type formulas plainly, as in C2H4.
The molecule contains a empirical units, so its molecular formula is b.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An analysis of the white ash left by a burnt ribbon gives the percentages below. Turn them into a ratio of atoms: divide each percentage by the element's relative atomic mass, to two decimal places, and then reduce the two results to the simplest whole-number ratio.
| percentage by mass | percentage divided by the atomic mass | simplest whole-number ratio | |
|---|---|---|---|
| $\mathrm{Mg}$ | 60 | ||
| $\mathrm{O}$ | 40 |
You can go from percentages to a formula and, with a molar mass, to the molecular formula. Say why ethyne and benzene have the same percentage composition, and what one extra measurement tells them apart. Next: using a balanced equation to get from the mass of one substance to the mass of another.
16. Your turn: 60.0% magnesium, 40.0% oxygen., step 3
$\mathrm{MgO}$
A ratio of 1 is not written.