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A sample of iron oxide was found to contain 1.116 g of iron and 0.480 g of oxygen. Its molar mass is roughly 5 x as great as that of oxygen gas. How do I find the empirical formula and the molecular formula of this compound?
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+ Inorganic chemistry
+ Iron oxide
+ Science
+ Chemistry
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Misaiel Dominguez
A sample of iron oxide was found to contain 1.116 g of iron and 0.480 g of oxygen. Its molar mass is roughly 5 x as great as that of oxygen gas. How do I find the empirical formula and the molecular formula of this compound?
This is an ionic compound, so you will be finding the empirical formula and the compound formula.
Convert both grams to moles.
Divide by the smaller.
If not both whole numbers, multiply BOTH by the number that will convert the decimal to a whole number( look at decimal equivalent of fractions if necessary.
That gives you the empirical formula.
Calculate the molar mass of O2 and multiply by 5 to get compounds molar mass. ( Oxygen gas is O2).
Calculate the mass of the empirical formula.
Divide the molar mass by the mass of the empiricsl formula. You should get a whole number. If not , make it a whole number by multiplying to get rid if decimal.
Multiply all the subscripts in the empirical formula by the whole number to get the actual formula.
This is an ionic compound, so you will be finding the empirical formula and the compound formula.
Convert both grams to moles.
Divide by the smaller.
If not both whole numbers, multiply BOTH by the number that will convert the decimal to a whole number( look at decimal equivalent of fractions if necessary.
That gives you the empirical formula.
Calculate the molar mass of O2 and multiply by 5 to get compounds molar mass. ( Oxygen gas is O2).
Calculate the mass of the empirical formula.
Divide the molar mass by the mass of the empiricsl formula. You should get a whole number. If not , make it a whole number by multiplying to get rid if decimal.
Multiply all the subscripts in the empirical formula by the whole number to get the actual formula.
A sample of iron oxide was found to contain 1.116 g of iron and 0.480 g of oxygen. Its molar mass is roughly 5 x as great as that of oxygen gas. How do I find the empirical formula and the molecular formula of this compound?
First, convert the mass of Fe and Oxygen to moles.
Molar mass of Fe = 55.85 g/mol.
Mole of Fe in 1.116 g = (1.116 g/55.85 g/mol) = 0.02 mol
Molar mass of O = 16.0 g/mol.
Mole of O in 0.480 g = (0.48 g/16 g/mol) = 0.03 mol
Hence, the number of mole in each constituent as
Fe = 0.02 mol
O = 0.03 mol
7. Divide throughout by the smaller number:
Hence, Fe = 0.02/0.02 =1
O = 0.03/0.02 =1.5
Multiply each by 2 for whole number.
Then, Fe = 2 and O = 3
8. Hence, the empirical formula of iron oxide = (Fe2O3)n
9. Its molar mass = 5 x 32 g/mol = 160 g/mol.
10. Therefore, from step 8 and 9, (Fe2O3)n = 160.
11. That is, (2 x 55.85 + 3 x 16)n = 160,
(111.7 + 48)n = 160. On solving for n = 160/159.7 = 1
12. Hence, the empirical formula of iron oxide = (Fe2O3)1 = Fe2O3.
13. From the above calculations, the molecular formula of iron oxide is the same as its empirical formula.
14. Therefore, the molecular formula of Iron oxide is Fe2O3.
A sample of iron oxide was found to contain 1.116 g of iron and 0.480 g of oxygen. Its molar mass is roughly 5 x as great as that of oxygen gas. How do I find the empirical formula and the molecular formula of this compound?
First, convert the mass of Fe and Oxygen to moles.
Molar mass of Fe = 55.85 g/mol.
Mole of Fe in 1.116 g = (1.116 g/55.85 g/mol) = 0.02 mol
Molar mass of O = 16.0 g/mol.
Mole of O in 0.480 g = (0.48 g/16 g/mol) = 0.03 mol
Hence, the number of mole in each constituent as
Fe = 0.02 mol
O = 0.03 mol
7. Divide throughout by the smaller number:
Hence, Fe = 0.02/0.02 =1
O = 0.03/0.02 =1.5
Multiply each by 2 for whole number.
Then, Fe = 2 and O = 3
8. Hence, the empirical formula of iron oxide = (Fe2O3)n
9. Its molar mass = 5 x 32 g/mol = 160 g/mol.
10. Therefore, from step 8 and 9, (Fe2O3)n = 160.
11. That is, (2 x 55.85 + 3 x 16)n = 160,
(111.7 + 48)n = 160. On solving for n = 160/159.7 = 1
12. Hence, the empirical formula of iron oxide = (Fe2O3)1 = Fe2O3.
13. From the above calculations, the molecular formula of iron oxide is the same as its empirical formula.
14. Therefore, the molecular formula of Iron oxide is Fe2O3.
This is an ionic compound, so you will be finding the empirical formula and the compound formula.
Convert both grams to moles.
Divide by the smaller.
If not both whole numbers, multiply BOTH by the number that will convert the decimal to a whole number( look at decimal equivalent of fractions if necessary.
That gives you the empirical formula.
Calculate the molar mass of O2 and multiply by 5 to get compounds molar mass. ( Oxygen gas is O2).
Calculate the mass of the empirical formula.
Divide the molar mass by the mass of the empiricsl formula. You should get a whole number. If not , make it a whole number by multiplying to get rid if decimal.
Multiply all the subscripts in the empirical formula by the whole number to get the actual formula.
This is an ionic compound, so you will be finding the empirical formula and the compound formula.
Convert both grams to moles.
Divide by the smaller.
If not both whole numbers, multiply BOTH by the number that will convert the decimal to a whole number( look at decimal equivalent of fractions if necessary.
That gives you the empirical formula.
Calculate the molar mass of O2 and multiply by 5 to get compounds molar mass. ( Oxygen gas is O2).
Calculate the mass of the empirical formula.
Divide the molar mass by the mass of the empiricsl formula. You should get a whole number. If not , make it a whole number by multiplying to get rid if decimal.
Multiply all the subscripts in the empirical formula by the whole number to get the actual formula.
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A sample of iron oxide was found to contain 1.116 g of iron and 0.480 g of oxygen. Its molar mass is roughly 5 x as great as that of oxygen gas. How do I find the empirical formula and the molecular formula of this compound?
Fe = 0.02 mol
O = 0.03 mol
7. Divide throughout by the smaller number:
Hence, Fe = 0.02/0.02 =1
O = 0.03/0.02 =1.5
Multiply each by 2 for whole number.
Then, Fe = 2 and O = 3
8. Hence, the empirical formula of iron oxide = (Fe2O3)n
9. Its molar mass = 5 x 32 g/mol = 160 g/mol.
10. Therefore, from step 8 and 9, (Fe2O3)n = 160.
11. That is, (2 x 55.85 + 3 x 16)n = 160,
(111.7 + 48)n = 160. On solving for n = 160/159.7 = 1
12. Hence, the empirical formula of iron oxide = (Fe2O3)1 = Fe2O3.
13. From the above calculations, the molecular formula of iron oxide is the same as its empirical formula.
14. Therefore, the molecular formula of Iron oxide is Fe2O3.
A sample of iron oxide was found to contain 1.116 g of iron and 0.480 g of oxygen. Its molar mass is roughly 5 x as great as that of oxygen gas. How do I find the empirical formula and the molecular formula of this compound?
Fe = 0.02 mol
O = 0.03 mol
7. Divide throughout by the smaller number:
Hence, Fe = 0.02/0.02 =1
O = 0.03/0.02 =1.5
Multiply each by 2 for whole number.
Then, Fe = 2 and O = 3
8. Hence, the empirical formula of iron oxide = (Fe2O3)n
9. Its molar mass = 5 x 32 g/mol = 160 g/mol.
10. Therefore, from step 8 and 9, (Fe2O3)n = 160.
11. That is, (2 x 55.85 + 3 x 16)n = 160,
(111.7 + 48)n = 160. On solving for n = 160/159.7 = 1
12. Hence, the empirical formula of iron oxide = (Fe2O3)1 = Fe2O3.
13. From the above calculations, the molecular formula of iron oxide is the same as its empirical formula.
14. Therefore, the molecular formula of Iron oxide is Fe2O3.
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