Did you know that you can use the empirical formula on organic compounds?
Now what is an organic compound?
An organic compound is any member of a large class of gaseous, liquid, or solid chemical compound whose molecules contain carbon.
Ex. When a 3.79 grams of an organic compound is burned, 6.61 grams of CO2 and 3.59 grams of H2O are produced. What is the empirical formula of this compound?
Molar mass of CO2: 12 + 16 + 16 = 44 g/mol
Mole C = 6.61g x 1 mole CO2 x 1 mole of C = 0.15023 mole of C
44.0 g CO2 1 mole of CO2
Molar mass of H20: 1.0 + 1.0 + 16.0 = 18.0 g/mol
Mole H = 3.59 g x 1 mole of H2O x 2 moles of H = 0.39889 mole of H
18.0 g H2O 1 mole of H20
Check the mass of C and H
Mass of C = 0.15023 x 12.0 g C = 1.80 g C
1 mole C
Mass of H = 0.39889 x 1.0 g H = 0.39889 g H
1 mole H
Since the masses of C and H do not add up to 3.79, the rest of the mass must be from O.
Mass of O = 3.79 - 1.80 - 0.39889 = 1.59111
Mole O = 1.59 x 1 mole O = 0.0994 g O
16.0 g O
Now we divide by the smallest molar amount. That is 0.0994
Carbon: 0.15023/0.0994 = 1.5 x 2 = 3
Hydrogen: 0.39889/0.0994 = 4.0 x 2 = 8
Oxygen: 0.0994/0.0994 = 1 x 2 = 2
Empirical formula: C2H8O2
This video will explain determining the empirical formula of an organic compound:
Empirical and Molecular Fomula
What is the empirical formula?
It gives the lowest term ratio of atoms (or moles) in a formula.
Ex. C4H10 (butane) --> molecular fomula
Reduce the subscripts to lowest terms to get the empirical formula
C2H5 --> empirical formula
Ex. Consider that we have 10.87 grams of Fe and 4.66 g of O. What is the empirical formula?
First, we must convert the grams to moles.
10.87g x 1mole = 0.195 moles of Fe
55.8g
4.66g x 1 mole = 0.291 moles of O
16g
Now, we divide the mole amount by the smallest mole which is 0.195
Fe: 0.195 = 1
0.195
O: 0.291 = 1.49 = 1.5
0.195
Next, we must get those two numbers to the smallest whole number
Fe 1 x 2 = 2
O 1.5 x 2 = 3
The empirical formula is: Fe2O3
Ex. A compound contains 31.9% of K, 28.9% of Cl, and 39.2% of O. What is the empirical formula?
*assume we have 100g*
31.9 x 1 mol = 0.816 moles of K
39.1g
28.9 x 1 mol = 0.814 moles of Cl
35.5g
39.2 x 1 mol = 2.45 moles of O
16g
lowest mole: 0.814
0.816 = 1.00
0.814
0.814 = 1
0.814
2.45 = 3
0.814
Empirical formula: KClO3
What is the molecular formula?
It is a multiple of the empirical formula and shows the actual numbers of atoms that combine to form a molecule.
To calculate the multiple, we use this formula:
n = molar mass of the compound
molar mass of the empirical formula
Ex. A molecule has an empirical formula of C2H5 and the molar mass is 58 g/mol. What is the molecular formula?
Molar mass is C2H5 = 58 g/mol
Molar mass of the empirical formula (total molar mass of all elements in the compound):
C2 = 12 x 2 = 24
H5 = 1 x 5 = 5
C2H5 = 29 g/mol
58 g/mol = 2
29 g/mol
2 x C2H5 = C4H10
Video about the empirical formula:
Video about the molecular formula:
It gives the lowest term ratio of atoms (or moles) in a formula.
Ex. C4H10 (butane) --> molecular fomula
Reduce the subscripts to lowest terms to get the empirical formula
C2H5 --> empirical formula
Ex. Consider that we have 10.87 grams of Fe and 4.66 g of O. What is the empirical formula?
First, we must convert the grams to moles.
10.87g x 1mole = 0.195 moles of Fe
55.8g
4.66g x 1 mole = 0.291 moles of O
16g
Now, we divide the mole amount by the smallest mole which is 0.195
Fe: 0.195 = 1
0.195
O: 0.291 = 1.49 = 1.5
0.195
Next, we must get those two numbers to the smallest whole number
Fe 1 x 2 = 2
O 1.5 x 2 = 3
The empirical formula is: Fe2O3
Ex. A compound contains 31.9% of K, 28.9% of Cl, and 39.2% of O. What is the empirical formula?
*assume we have 100g*
31.9 x 1 mol = 0.816 moles of K
39.1g
28.9 x 1 mol = 0.814 moles of Cl
35.5g
39.2 x 1 mol = 2.45 moles of O
16g
lowest mole: 0.814
0.816 = 1.00
0.814
0.814 = 1
0.814
2.45 = 3
0.814
Empirical formula: KClO3
What is the molecular formula?
It is a multiple of the empirical formula and shows the actual numbers of atoms that combine to form a molecule.
To calculate the multiple, we use this formula:
n = molar mass of the compound
molar mass of the empirical formula
Ex. A molecule has an empirical formula of C2H5 and the molar mass is 58 g/mol. What is the molecular formula?
Molar mass is C2H5 = 58 g/mol
Molar mass of the empirical formula (total molar mass of all elements in the compound):
C2 = 12 x 2 = 24
H5 = 1 x 5 = 5
C2H5 = 29 g/mol
58 g/mol = 2
29 g/mol
2 x C2H5 = C4H10
Video about the empirical formula:
Video about the molecular formula:
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