Empirical formula from percentages
CH₂Clfrom 4.07% H, 24.27% C, 71.65% Cl
Molar mass
Molar mass. The mass of one mole in grams: add the atomic mass of every atom in the formula. Masses. Standard atomic masses to two decimals, hydrogen 1.008, every element to uranium. Typing. Ca(OH)2, (NH4)2SO4, CuSO4.5H2O and K4[Fe(CN)6] all work.
18.02g/mol
H₂O
- H× 211.19%
- O× 188.81%
- H2 × 1.0082.016
- O1 × 16.0016.00
- Total18.02 g/mol
Grams, moles, particles
Type any one. The other three fill in. Moles. From grams, particles or litres they are rounded to 3 significant figures first, as in a worked answer. 1 mol = 6.022 × 10²³ particles = 22.7 L of gas at STP (1 bar); 22.4 L at 1 atm.
0.5 mol of H₂O = 3.011 × 10²³ molecules
At 1 atm (older value): 11.20 L
Empirical formula
1. Take 100 g, so each percentage is grams. 2. Divide by the atomic mass to get moles, then by the smallest. 3. Multiply by 2, 3 or 4 if a ratio ends in .5, .33 or .25. Molecular formula. Add the molar mass: n = molar mass ÷ empirical formula mass.
%
%
%
Empirical formulaCH₂Cl49.48 g/mol
| Element | g in 100 g | ÷ mass | ÷ least | Whole |
|---|---|---|---|---|
| H | 4.07 | 4.038 | 2.00 | 2 |
| C | 24.27 | 2.021 | 1.00 | 1 |
| Cl | 71.65 | 2.021 | 1.00 | 1 |
Limiting reagent
1. The equation is balanced for you. 2. Each mass becomes moles, divided by its coefficient. 3. The smallest runs out first: it is the limiting reagent and sets how much product forms. Leave a mass empty if that reactant is in excess.
N₂ + 3 H₂ → 2 NH₃
- N₂1784 mol ÷ 1 = 17843.67 kg left over
- H₂4960 mol ÷ 3 = 1653Limiting
Formed
- NH₃3307 mol56.3 kg
The NCERT way
percent ÷ atomic mass, ÷ smallest, then make whole numbers
A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Find its empirical formula.
- In 100 g there are 4.07 g of H, 24.27 g of C, 71.65 g of Cl.
- Divide by the atomic masses: H 4.07 ÷ 1.01 = 4.038, C 24.27 ÷ 12.01 = 2.021, Cl 71.65 ÷ 35.45 = 2.021.
- Divide by the smallest, 2.021: H 2.00, C 1.00, Cl 1.00, which round to 2, 1, 1.
- Empirical formula: CH₂Cl, with a mass of 49.48 g/mol.
- The empirical formula is CH₂Cl.
Empirical formula: quick answers
How do you find an empirical formula from percentages?
Take 100 g, so each percentage becomes grams. Divide each by its atomic mass to get moles, divide all of them by the smallest, and multiply if needed to get small whole numbers.
What if the ratio is 1.5 or 1.33?
Multiply every ratio by the same small number to make them whole: by 2 for 1.5, by 3 for 1.33 or 1.67, and by 4 for 1.25.
What is the empirical formula of glucose?
Glucose is C₆H₁₂O₆. Dividing every subscript by 6 gives CH₂O, the simplest whole-number ratio.