Answer :
To determine the empirical formula of the compound, we assume a basis of 100 g of this compound. We calculate as follows:
C = 68.75 g
H = 10.90 g
O = 20.35 g
We convert these mass to moles,
C = 68.75 g / 12.01 g/mol = 5.72 mol
H = 10.90 g / 1.01 g/mol = 10.79 mol
O = 20.35 g / 16 g/mol = 1.27 mol
C = 5.72 mol / 1.27 mol = 5
H = 10.79 mol / 1.27 mol = 8
O = 1.27 mol / 1.27 mol = 1
C5H8O
C = 68.75 g
H = 10.90 g
O = 20.35 g
We convert these mass to moles,
C = 68.75 g / 12.01 g/mol = 5.72 mol
H = 10.90 g / 1.01 g/mol = 10.79 mol
O = 20.35 g / 16 g/mol = 1.27 mol
C = 5.72 mol / 1.27 mol = 5
H = 10.79 mol / 1.27 mol = 8
O = 1.27 mol / 1.27 mol = 1
C5H8O
Answer: The empirical formula is [tex]C_9H_{17}O_2[/tex]
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C= 68.75 g
Mass of H = 10.90 g
Mass of O = 20.35 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{68.75g}{12g/mole}=5.73moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{10.90g}{1g/mole}=10.90moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{20.35g}{16g/mole}=1.27moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{5.73}{1.27}=4.5[/tex]
For H = [tex]\frac{10.90}{1.27}=8.6[/tex]
For O =[tex]\frac{1.27}{1.27}=1[/tex]
The ratio of C : H: O= 4.5: 8.6: 1
converting them into whole number ratio by multiplying by 2
[tex]2\times C_{4.5}H_{8.6}O=C_9H_{17}O_2[/tex]
Hence the empirical formula is [tex]C_9H_{17}O_2[/tex]