Molecular formula from the empirical formula
C₂H₄Cl₂from CH₂Cl and 98.96 g/mol
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
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
molecular formula = (empirical formula) × n, where n = molar mass ÷ empirical formula mass
A compound has the empirical formula found from 4.07% H, 24.27% C and 71.65% Cl, and a molar mass of 98.96 g/mol. Find its molecular 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.
- n = molar mass ÷ empirical mass = 98.96 ÷ 49.48 = 2.
- Molecular formula: (CH₂Cl) × 2 = C₂H₄Cl₂.
- The molecular formula is C₂H₄Cl₂.
Molecular formula: quick answers
How do you find the molecular formula?
Work out the empirical formula and its mass. Divide the molar mass by that mass to get a whole number n, then multiply every subscript by n.
Can the molecular formula be the same as the empirical formula?
Yes, when n = 1. Water, H₂O, and carbon dioxide, CO₂, are examples. Glucose, C₆H₁₂O₆, has the empirical formula CH₂O, so n = 6.
What is the molecular formula of CH₂Cl with a molar mass of 98.96 g/mol?
The empirical formula mass is 49.48 g/mol, so n = 98.96 ÷ 49.48 = 2 and the molecular formula is C₂H₄Cl₂.