Respuesta :
According to law of definite proportion:
In a compound, elements are always arranged in fixed ratio by mass.
Here, sample 1 has 23.22 g Carbon and 32.00 g Oxygen.
Converting mass into number of moles:
Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,
[tex]n_{C}=\frac{m_{C}}{M_{C}}=\frac{24.22 g}{12 g/mol}\approx 2 mol[/tex]
Similarly, number of moles of oxygen will be:
[tex]n_{O}=\frac{m_{O}}{M_{O}}=\frac{32 g}{16 g/mol}=2 mol[/tex]
The ratio of number of moles of carbon and oxygen will be:
[tex]C:O=n_{C}:n_{O}=2:2=1:1[/tex]
Therefore, formula of compound will be CO.
Sample 2:
It has 36.22 g Carbon and 48.00 g Oxygen.
Converting mass into number of moles:
Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,
[tex]n_{C}=\frac{m_{C}}{M_{C}}=\frac{36.22 g}{12 g/mol}\approx 3 mol[/tex]
Similarly, number of moles of oxygen will be:
[tex]n_{O}=\frac{m_{O}}{M_{O}}=\frac{48 g}{16 g/mol}=3 mol[/tex]
The ratio of number of moles of carbon and oxygen will be:
[tex]C:O=n_{C}:n_{O}=3:3=1:1[/tex]
The formula of compound will be CO.
Therefore, it is proved that carbon and oxygen are present in fixed ratios in both the samples.
