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,)

1. How many moles are there in 5.0 g of methanol (CH3OH)?
How many molecules of CuCO3 would weigh 9.0 g?
How many moles would 3.0 g of cyclohexane (C6H12) contain?
How much would 4.00 moles of benzene weigh?
What is the mass of 1000 atoms of calcium?

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 * 10^(23))=\boxed{4.4 * 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
`(1.000)/(6.022 * 10^(23))` moles.

This is equal to
`(1000)/(6.022 * 10^(23)) \cdot 40.078=\boxed{6.7 * 10^(-20)}` grams.