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Sagot :
When a substance has molar mass of 60g/mol with 40 percent carbon, 6.7 percent hydrogen and 53.3 percent oxygen, the empirical formula is CH₂O.
given,
molar mass of the substance= 60 g/mol
percentage of C= 40%
percentage of H= 6.7%
percentage of O= 53.3%
we know that,
molar mass of C= 12g/mol
molar mass of H = 1g/mol
molar mass of O= 16g/mol
The ratio of C,H and O is
[tex]\frac{40g}{12g/mol} C : \frac{6.7g}{1g/mol} H : \frac{53.5g}{16g/mol} O[/tex]
After dividing each, we get the ratio
C:H:O = 3.33 : 6.65 : 3.34
Now, we will divide each elemental ratio by the lowest whole number i.e 3.33 to get the simplest whole number ratio.
C = 3.33/3.33 =1
H = 6.65/3.33 = 1.99≈ 2
O = 3.34/3.33 =1
therefore, the required simplest whole number ratio is C:H:O = 1:2:1
Thus, the empirical formula is CH₂O.
To know more about empirical formula here
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