What is the empirical formula of a compound that is 14g calcium, 11g oxygen, and 0.7g hydrogen?
A. CaOH
B. Ca(OH)2
C. Ca2OH2
D. CaO2H
Convert g to mols.
Ca = 14/atomic mass Ca.
O = 11/atomic mass O.
H = 0.7/1 = 0.7
Then find the ratio of the elements to each other. The easy way to get started on that is to divide the smallest number by itself then follow suit with the other numbers. Don't round too much.
To determine the empirical formula of a compound, you need to find the simplest whole number ratio of the elements present in the compound.
Step 1: Convert the given masses of the elements to moles.
- For calcium (Ca):
- Moles of Ca = Mass of Ca (14g) / Molar mass of Ca (40.08 g/mol) ≈ 0.349 mol
- For oxygen (O):
- Moles of O = Mass of O (11g) / Molar mass of O (16.00 g/mol) ≈ 0.688 mol
- For hydrogen (H):
- Moles of H = Mass of H (0.7g) / Molar mass of H (1.01 g/mol) ≈ 0.693 mol
Step 2: Divide the number of moles of each element by the smallest number of moles to get the simplest ratio.
- The smallest number of moles is approximately 0.349 mol (calcium).
- Dividing the number of moles for each element by 0.349 mol gives you:
- Moles of Ca = 0.349 mol / 0.349 mol = 1
- Moles of O = 0.688 mol / 0.349 mol ≈ 1.97
- Moles of H = 0.693 mol / 0.349 mol ≈ 1.99
Step 3: Round off the moles to the nearest whole number to get the subscripts in the empirical formula.
- The empirical formula will be CaOH2 (rounded from Ca1O1.97H1.99).
Therefore, the empirical formula of the compound is B. Ca(OH)2.