Question

(Atomic masses used are C = 12.0 g mol–1, O = 16.0 g mol–1, H = 1.01 g mol–1, Cu = 63.5 g mol–1,)

Answer 1

Question 1

The formula mass of methanol is $12.0+3(1.01)+16.0+1.01=32.04$ g/mol.

Therefore, in 5.0 grams of methanol, there are $5.0/32.04=\boxed{0.16}$ moles.

Question 2

The formula mass of copper(II) carbonate is $63.5+12.0+3(16.0)=123.5$ g/mol.

This means that in 9.0 grams of copper(II) carbonate, there are $9.0/123.5=0.072874493927126$ moles.

Using Avogradro's number, there are $(0.072874493927126)(6.022 \times 10^{23})=\boxed{4.4 \times 10^{22}}$ molecules.

Question 3

The formula mass of cyclohexane is $12.0(6)+1.01(12)=84.12$ g/mol.

So, 3.0 grams of cyclohexane is the same as $3.0/84.12=\boxed{0.036}$ moles.

Question 4

The formula mass of benzene is $12.0(6)+1.01(6)=78.06$ g/mol.

This means that 4.00 moles of benzene has a mass of $(78.06)(4.00)=\boxed{312}$ grams.

Question 5

1000 atoms of calcium is the same as $\frac{1.000}{6.022 \times 10^{23}}$ moles.

This is equal to $\frac{1000}{6.022 \times 10^{23}} \cdot 40.078=\boxed{6.7 \times 10^{-20}}$ grams.