Determine the empirical formula of each compound from the given weights:
1) 7.615 g Ga, 2.622 g O
2) O 11.89 g Fe, 5.11 g O
3) 0.366 g Na, 0.220 g N, 0.752 g O
4) 87.3 g Na, 121.5 g S, 91.2 g O
1) 7.615 g Ga, 2.622 g O
2) O 11.89 g Fe, 5.11 g O
3) 0.366 g Na, 0.220 g N, 0.752 g O
4) 87.3 g Na, 121.5 g S, 91.2 g O
Add an Answer
More Questions From This User See All
Answers & Comments
1.
Ga: 7.615 g / 69.72 g/mol = 0.1093 mol
O: 2.622 g / 16.00 g/mol = 0.1639 mol
Divide by the smallest number of moles to get the simplest whole-number ratio:
Ga:O = 0.1093 mol / 0.1093 mol = 1
O:Ga = 0.1639 mol / 0.1093 mol = 1.5
Multiply by 2 to get the smallest whole-number ratio:
Ga2O3
2.
Fe: 11.89 g / 55.85 g/mol = 0.2129 mol
O: 5.11 g / 16.00 g/mol = 0.3194 mol
Divide by the smallest number of moles to get the simplest whole-number ratio:
Fe:O = 0.2129 mol / 0.2129 mol = 1
O:Fe = 0.3194 mol / 0.2129 mol = 1.5
Multiply by 2 to get the smallest whole-number ratio:
Fe2O3
3.
Na: 0.366 g / 22.99 g/mol = 0.0159 mol
N: 0.220 g / 14.01 g/mol = 0.0157 mol
O: 0.752 g / 16.00 g/mol = 0.047 mol
Divide by the smallest number of moles to get the simplest whole-number ratio:
Na:N:O = 0.0159 mol / 0.0157 mol / 0.0157 mol = 1:1:1
The empirical formula is NaNO.
4.
Na: 87.3 g / 22.99 g/mol = 3.799 mol
S: 121.5 g / 32.06 g/mol = 3.784 mol
O: 91.2 g / 16.00 g/mol = 5.7 mol
Divide by the smallest number of moles to get the simplest whole-number ratio:
Na:S:O = 3.799 mol / 3.784 mol / 5.7 mol = 1:1:3
The empirical formula is Na2SO3.
:)