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#empirical-formula
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A compound contains 40% Ca, 12%C and 48% O
find it's empirical formula and name the compound ?
Anonymous
Mar 22
A compound contains 40% Ca, 12%C and 48% O find it's empirical formula and name the compound ?
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ChatGPT
Mar 22
To find the empirical formula of the compound, we need to determine the ratio of each element present in the compound.
1. Convert the percentages to grams:
- 40% Ca = 40g Ca
- 12% C = 12g C
- 48% O = 48g O
2. Determine the moles of each element:
- Moles of Ca = 40g Ca / 40.08 g/mol = 0.998 mol Ca
- Moles of C = 12g C / 12.01 g/mol = 0.999 mol C
- Moles of O = 48g O / 16.00 g/mol = 3.00 mol O
3. Divide each mole value by the smallest mole value to get the simplest ratio:
- Ca: 0.998 / 0.998 = 1
- C: 0.999 / 0.998 = 1
- O: 3.00 / 0.998 = 3
Therefore, the empirical formula of the compound is CaCO3, which is calcium carbonate.
1. Convert the percentages to grams:
- 40% Ca = 40g Ca
- 12% C = 12g C
- 48% O = 48g O
2. Determine the moles of each element:
- Moles of Ca = 40g Ca / 40.08 g/mol = 0.998 mol Ca
- Moles of C = 12g C / 12.01 g/mol = 0.999 mol C
- Moles of O = 48g O / 16.00 g/mol = 3.00 mol O
3. Divide each mole value by the smallest mole value to get the simplest ratio:
- Ca: 0.998 / 0.998 = 1
- C: 0.999 / 0.998 = 1
- O: 3.00 / 0.998 = 3
Therefore, the empirical formula of the compound is CaCO3, which is calcium carbonate.
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