Answer :
This is an incomplete question, here is a complete question.
Oxalic acid is an organic substance. Its composition is 26.7% C, 2.2% H, and 71.1% O (by mass), and its molecular weight is 90 amu. What is its molecular formula?
Answer : The molecular formula of a compound is, [tex]C_2H_2O_4[/tex]
Solution : Given,
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 = 26.7 g
Mass of H = 2.2 g
Mass of O = 71.1 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{26.7g}{12g/mole}=2.225moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.2g}{1g/mole}=2.2moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{71.1g}{16g/mole}=4.444moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{2.225}{2.2}=1.0\aaprox 1[/tex]
For H = [tex]\frac{2.2}{2.2}=1[/tex]
For O = [tex]\frac{4.444}{2.2}=2.02\approx 2[/tex]
The ratio of C : H : O = 1 : 1 : 2
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_1H_1O_2=CHO_2[/tex]
The empirical formula weight = 1(12) + 1(1) + 2(16) = 45 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :
[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}[/tex]
[tex]n=\frac{90}{45}=2[/tex]
Molecular formula = [tex](C_1H_1O_2)_n=(C_1H_1O_2)_2=C_2H_2O_4[/tex]
Therefore, the molecular of the compound is, [tex]C_2H_2O_4[/tex]