chemical calculation

Question

a density of gold is 19.3 g/cm cube. How many atoms and how many moles of atoms does the volume equal to 1 mm cube of this metal contain?

Answer 1

The Atomic Mass of Gold (Au) is 196.96657 u. Which means 1 mole of Au weighs 196.96657 g. This question can be solved using dimensional analysis by first converting grams of Au per cm³ to grams of Au per mm³. Next, we will use the molecular weight to convert grams of Au to atoms and moles.

Remember that 1 m = 100 cm = 1000 mm. Therefore there are 10 mm in 1 cm. 1 cm³ can be expressed as 1 cm⋅cm⋅cm, and 1 mm³ as 1 mm⋅mm⋅mm. This allows us to use dimensional analysis to calculate cm³ to mm³ as follows:

(1 cm⋅cm⋅cm) × (10 mm⋅cm^-1) × (10 mm⋅cm^-1) × (10 mm⋅cm^-1) = 1000 mm³

Back to the question, we have

1 mm³ × (1 cm³ ÷ 1000 mm³) × (19.3 g Au ÷ 1 cm³) × (1 mol Au ÷ 196.96657 g Au) = 9.80 × 10^-5 mol Au

Avogadro's Number = 6.022 x 10²³ and the unit is mol^-1. All we have to do is multiply the number of moles of Au we have by Avogadro's Number to get the number of Au atoms.

# atoms of Au = (9.80 × 10^-5 mol Au) × (6.022 × 10²³ mol^-1) = 59.0 × 10¹⁸ = 5.90 × 10¹⁹ atoms