Percentage composition
52.14% Cin ethanol, C₂H₅OH
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.
46.07g/mol
C₂H₅OH
- C× 252.14%
- H× 613.13%
- O× 134.73%
- C2 × 12.0124.02
- H6 × 1.0086.048
- O1 × 16.0016.00
- Total46.07 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 C₂H₅OH = 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
Molecular formulaC₂H₄Cl₂n = 98.96 ÷ 49.48 = 2
| 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
mass % of an element = (mass of the element in 1 mol ÷ molar mass) × 100
Find the percentage of each element in ethanol, C₂H₅OH.
- Molar mass = 2 × 12.01 + 6 × 1.008 + 1 × 16.00 = 46.07 g/mol.
- Carbon: 24.02 ÷ 46.07 × 100 = 52.14%.
- Hydrogen: 6.048 ÷ 46.07 × 100 = 13.13%.
- Oxygen: 16.00 ÷ 46.07 × 100 = 34.73%.
- Ethanol is 52.14% carbon, 13.13% hydrogen, 34.73% oxygen.
Percentage composition: quick answers
How do you find the percentage composition?
Work out the molar mass. For each element, divide the mass of that element in one mole by the molar mass and multiply by 100. The percentages add up to 100.
What is the percentage of carbon in ethanol?
52.14%: 24.02 g of carbon in every 46.07 g of ethanol.
What is the use of percentage composition?
It tells you how pure a sample is, and it is the starting point for finding the empirical formula of a new compound.