The mole

Molar volume at STP

2.27 Lfor 4.4 g of CO₂ at STP

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.

Formula

44.01g/mol

CO₂

C× 127.29%
O× 272.71%

C1 × 12.0112.01
O2 × 16.0032.00
Total44.01 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.

Mass g Moles mol Particles molecules Gas at STP L

0.100 mol of CO₂ = 6.022 × 10²² molecules

At 1 atm (older value): 2.24 L

In mL of solution: 0.400 M

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.

%

Molar mass g/mol

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.

Equation

N₂kgH₂kg

N₂ + 3 H₂ → 2 NH₃

N₂1784 mol ÷ 1 = 17843.67 kg left over
H₂4960 mol ÷ 3 = 1653Limiting

Formed

NH₃3307 mol56.3 kg

Worked example

The NCERT way

volume at STP (L) = moles × 22.7

What volume does 4.4 g of CO₂ fill at STP?

Moles of CO₂ = 4.4 ÷ 44.01 = 0.100 mol.
Volume = moles × 22.7 L = 0.100 × 22.7 = 2.27 L at 1 bar.
With the older value at 1 atm: 0.100 × 22.4 = 2.24 L.
4.4 g of CO₂ fills 2.27 L at STP.

Questions