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a substance has a molar mass of 60 g/mol with 40.0 percent carbon, 6.7 percent hydrogen, and 53.3 percent oxygen. what is its empirical formula?

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.

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