Mole calculations working out 1

2 moles of HNO₃

H = 1, N = 14, O = 16

HNO₃ = 1 + 14 + (3 × 16) = 63 g mol^-1

2 moles of HNO₃ = 2 × 63 = 126 g

0.0312 moles of Cu(OH)₂

Cu = 63.5, O = 16, H = 1

Cu(OH)₂ = 63.5 + 2 × (16 + 1) = 97.5 g mol^-1

0.0312 moles of Cu(OH)₂ = 0.0312 × 97.5 = 3.042 g

Answer = 3.04 g (to 3 sig. figs.)

5.61 × 10^-4 moles of FeSO₄.7H₂O

H = 1, O =16

H₂O = (2 × 1) + 16 = 18 g mol^-1

Fe = 56, S = 32, O = 16, H₂O = 18

FeSO₄.7H₂O = 56 + 32 + (4 × 16) + (7 × 18) = 278 g mol^-1

5.61 × 10^-4 moles of FeSO₄.7H₂O = 5.61 × 10^-4 × 278 = 0.155958 g**

Answer = 0.156 g (to 3 sig. figs.)

**We cannot justify all these significant figures as we have used atomic molar masses which have been rounded off (usually to the nearest whole number).

1.7 moles of C₇H₆O₂

C = 12, H = 1, O = 16

C₇H₆O₂ = (7 × 12) + (6 × 1) + 2 × 16) = 122 g mol^-1

1.7 moles of C₇H₆O₂ = 1.7 × 122 = 207.4 g

Answer = 207 g (to 3 sig. figs.)

0.188 moles of ammonium dichromate

Ammonium dichromate has the formula (NH₄)₂Cr₂O₇ - if you didn't know this, check the section on ionic formulae.

N = 14, H =1, Cr = 52, O = 16

NH₄ = 14 + (4 × 1) = 18 g mol^-1

(NH₄)₂Cr₂O₇ = (2 × 18) + (2 × 52) + (7 × 16) = 252 g mol^-1

0.188 moles of (NH₄)₂Cr₂O₇ = 0.188 × 252 = 47.376 g

Answer = 47.4 g (to 3 sig. figs.)