Question

10.2 conversions between moles and atoms
1.) what is the mass of 5.119 x 10^23 molecules of copper sulfate (cuso₄)? reminder to use dimensional analysis from unit 4 to work out the problems. you will have to work them out this way on the test.
(1 point)
○ 159.6 g
○ 94.88 g
○ 0.85 g
○ 135.7 g
2.) how many moles of iron in a sample that contains 7.91 x 10^23 atoms of iron? (1 point)
○ 7.91 mol fe
○ 0.761 mol fe
○ 1.31 mol fe
○ 47.6 mol fe
3.) how many moles would 1.00 x 10^23 molecules of methane (ch₄) represent? (1 point)
○ 16.6 mol
○ 0.166 mol
○ 1.66 mol

Explanation:

Step1: Recall Avogadro's number

Avogadro's number ($N_A$) is $6.022\times 10^{23}$ molecules/mol.

Step2: Calculate moles for each problem

Problem 1

First, find the number of moles of $CuSO_4$. The number of moles $n=\frac{5.119\times 10^{23}\text{ molecules}}{6.022\times 10^{23}\text{ molecules/mol}}\approx0.85\text{ mol}$. The molar - mass of $CuSO_4$: $M_{CuSO_4}=63.55 + 32.07+4\times16.00 = 159.62\text{ g/mol}$. Then the mass $m = n\times M=0.85\text{ mol}\times159.62\text{ g/mol}\approx135.7\text{ g}$.

Problem 2

The number of moles of Fe, $n=\frac{7.91\times 10^{23}\text{ atoms}}{6.022\times 10^{23}\text{ atoms/mol}}\approx1.31\text{ mol}$.

Problem 3

The number of moles of $CH_4$, $n=\frac{1.00\times 10^{23}\text{ molecules}}{6.022\times 10^{23}\text{ molecules/mol}}\approx0.166\text{ mol}$.

Answer:

1. D. 135.7 g
2. C. 1.31 mol Fe
3. B. 0.166 mol