# Question 1cb75

Jun 15, 2014

He was a little careless with significant figures at that point.

The calculation should have been

${n}_{1}$ = 50.0 g O₂ × $\text{1 mol O₂"/"32.00 g O₂}$ = 1.5625 mol O₂ = 1.56 mol O₂

The mass of O₂ determines the number of significant figures.

The calculation for ${n}_{2}$ is

${n}_{2}$ = (V_2 × n_1)/V_1 = "79 L × 1.5625 mol"/"48 L"# = 2.5716… mol = 2.6 mol

You use all the digits in the moles of O₂, but you must round the answer to two significant figures, because the volumes have only two significant figures.

When you calculate the mass of O₂, you again use all the digits that your calculator gave for the moles. Then you round off the final answer to two significant figures.

Mass of O₂ = $\text{2.5716… mol O₂ × 32.00 g O₂"/"1mol O₂}$ = 82.2916… g O₂ = 82 g O₂

He multiplied by the rounded number of moles and got 83 g of O₂. This is called round-off error.

Hope this helps.