Percent Composition and Empirical and Molecular Formulas
A bag of lawn fertilizer at the hardware store has three numbers on the front, like 46-0-0. The first number tells you how much nitrogen is in the bag. But 46 of what? Pounds? Percent? And how does a chemist know the formula of a compound she has never seen before? Both questions have the same answer: mass data. You will come back to the fertilizer at the end of this page.
Percent composition by mass
Every compound has a fixed recipe. Water is always $\text{H}_2\text{O}$, so in any sample of water, the same fraction of the mass is hydrogen. That fraction, written as a percent, is the percent composition by mass.
To find it, you pretend you have exactly 1 mole of the compound. Then the mass of each element in that mole comes straight from the formula and the periodic table.
Why 1 mole? Because the formula counts atoms, and a mole is the bridge between atoms and grams. The mass of an element in 1 mole is (number of atoms in the formula) × (its atomic mass in g/mol). The percent does not depend on the sample size, so any amount of the compound gives the same percent.
Define percent composition by mass* of a compound. Then write the expression you would use to calculate the percent by mass of oxygen in a compound, using the terms…
Define percent composition by mass of a compound. Then write the expression you would use to calculate the percent by mass of oxygen in a compound, using the terms "mass of oxygen in one mole of compound" and "molar mass of the compound."
Calculate the percent composition by mass of hydrogen and oxygen in water, $\text{H}2\text{O}$. Give your answers to three significant figures.
Calculate the percent composition by mass of hydrogen and oxygen in water, $\text{H}_2\text{O}$. Give your answers to three significant figures.
Aaliyah is studying fertilizers. Calculate the percent by mass of nitrogen in ammonium nitrate, $\text{NH}4\text{NO}3$, and in urea, $\text{CO(NH}2)2$. Which compound is…
Aaliyah is studying fertilizers. Calculate the percent by mass of nitrogen in ammonium nitrate, $\text{NH}_4\text{NO}_3$, and in urea, $\text{CO(NH}_2)_2$. Which compound is the richer source of nitrogen per gram? Show your work.
From percent composition to an empirical formula
Now run the process backward. A lab can measure how many grams of each element are in a sample, but nobody can see the atoms. You need to turn the masses into a formula.
The formula tells you the ratio of atoms, and atoms are counted in moles. So the plan is: grams → moles → simplest whole-number ratio.
If the problem gives percents, assume a 100.0 g sample. Then each percent becomes the same number of grams (43.6% → 43.6 g). If it gives grams, use them directly.
In the lab, Mateo analyzes a 2.500 g sample of a compound and finds that it contains 1.000 g of carbon, 0.168 g of hydrogen, and the rest is oxygen. Find the mass of…
In the lab, Mateo analyzes a 2.500 g sample of a compound and finds that it contains 1.000 g of carbon, 0.168 g of hydrogen, and the rest is oxygen. Find the mass of oxygen, convert each mass to moles, and determine the empirical formula of the compound.
Priya analyzes an unknown iron oxide and finds that it is 69.9% iron by mass, the rest being oxygen. (a) Determine the empirical formula of the oxide. (b) Calculate the…
Priya analyzes an unknown iron oxide and finds that it is 69.9% iron by mass, the rest being oxygen. (a) Determine the empirical formula of the oxide. (b) Calculate the percent iron by mass in $\text{FeO}$ and in $\text{Fe}_2\text{O}_3$, and use your results to argue which of these two compounds Priya has. (c) Explain why a 100.0 g sample is a convenient assumption for this kind of problem.
Empirical vs. molecular formulas
Glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) and formaldehyde ($\text{CH}_2\text{O}$) are very different substances, but both reduce to the ratio C : H : O = 1 : 2 : 1. The empirical formula only gives the ratio. It cannot tell you how many atoms are in one actual molecule.
To find the multiplier, you need one extra piece of data: the molar mass of the real compound (measured in the lab).
Why does this work? The molecule is $n$ copies of the empirical unit, so its mass is $n$ times the mass of that unit.
Classify each formula as an empirical formula only, a molecular formula only, or possibly both: $\text{C}6\text{H}{12}\text{O}6$, $\text{CH}2\text{O}$,…
Classify each formula as an empirical formula only, a molecular formula only, or possibly both: $\text{C}_6\text{H}_{12}\text{O}_6$, $\text{CH}_2\text{O}$, $\text{H}_2\text{O}$, $\text{C}_2\text{H}_4$. Give a one-sentence reason for each.
A compound has the empirical formula $\text{CH}2\text{O}$ and a molar mass of 180.16 g/mol. Determine its molecular formula. Then explain why the empirical formula alone…
A compound has the empirical formula $\text{CH}_2\text{O}$ and a molar mass of 180.16 g/mol. Determine its molecular formula. Then explain why the empirical formula alone cannot tell you the molar mass of a compound.
A compound is analyzed by combustion. A 1.000 g sample is found to contain 0.5217 g of carbon, 0.1304 g of hydrogen, and 0.3479 g of oxygen. Its molar mass is determined…
A compound is analyzed by combustion. A 1.000 g sample is found to contain 0.5217 g of carbon, 0.1304 g of hydrogen, and 0.3479 g of oxygen. Its molar mass is determined to be about 46 g/mol. (a) Find the empirical formula. (b) Find the molecular formula. (c) Another compound has the same empirical formula but a molar mass near 92 g/mol. Write its molecular formula and explain whether its percent composition by mass differs from that of the first compound.
Summary
Back to the fertilizer bag: the first number in 46-0-0 is the percent nitrogen by mass. It comes from the same percent composition calculation you just learned. Dividing the mass of nitrogen in one mole of a compound by its molar mass tells you how rich a source of nitrogen it is.