Notes: Mol ratios
In a balanced chemical equation the mole ratio is determined by the coefficients.
Coefficients = # of moles
2 HgO  2 Hg + O2
2 moles of HgO breaks down to produce 2 moles of Hg and 1 mole of O2.
Mol ratio:
2:2:1 HgO:Hg:O2
2:2 HgO:Hg
2:1 HgO:O2
2:1 Hg:O2
grams = mol ∙ g/mol
mol = grams ÷ g/mol
Calculating molecular weights:
Hg = 200.59 AMU = 201 AMU = 201 grams / mole [Hg]
O = 15.9994 AMU = 16 AMU = 16 grams / mole [O]
O2 = 2(16 grams / mole) = 32 grams / mole [O2]
HgO = 201 AMU + 16 AMU = 217 AMU = 217 grams / mole [HgO]
Sample problem 1:
How many grams of HgO are required to produce 96 grams of O2?
96 grams of O2 = 96 grams ÷ 32 grams / mole [O2] = 3 moles of O2
HgO:O2 = 2:1 ratio
6 moles of HgO : 3 moles of O2
6 moles [HgO] • 217 grams / mole [HgO] = 1302 grams HgO
Sample problem 2:
How many moles of O2 will be produced by 6 moles of HgO?
HgO:O2 = 2:1 ratio
6 moles of HgO : 3 moles of O2
Sample problem 3:
How many grams of HgO will be required to produce 0.5 mol of Hg?
HgO:Hg = 2:2 ratio
0.5 moles [HgO] : 0.5 moles [Hg]
0.5 moles • 217 grams / mole [HgO] = 108.5 grams [HgO]