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Sagot :
Answer: The molecular formula will be [tex]C_2H_4O_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 = 40.0 g
Mass of O = 53.3 g
Mass of H = 6.66 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40.0g}{12g/mole}=3.33moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.3g}{16g/mole}=3.33moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.66g}{1g/mole}=6.66moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{3.33}{3.33}=1[/tex]
For O =[tex]\frac{3.33}{3.33}=1[/tex]
For H = [tex]\frac{6.66}{3.33}=2[/tex]
The ratio of C : O : H = 1: 1: 2
Hence the empirical formula is [tex]COH_2[/tex]
The empirical weight of [tex]COH_2[/tex] = 1(12)+1(16)+2(1)= 30g.
The molecular weight = 60 g/mole
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{60}{30}=2[/tex]
The molecular formula will be=[tex]2\times CH_2O=C_2H_4O_2[/tex]
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