The empirical formula of a compound is $C_{2} H_{4} O$ . If the empirical formula mass is equal to one half of its vapor density, find the gram molecular mass of the compound.

176

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Solution

The correct option is B 176
Emperical formula mass of $C_{2} H_{4} O$ = 44g
Given that vapour density = $2 \times 44$ = 88g
Molecular mass = 2(vapour density)
Therefore molecular mass = $2 \times 88$ = 176g

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The empirical formula of a compound is C2H4O. If the empirical formula mass is equal to one half of its vapor density, find the gram molecular mass of the compound.

The correct option is B 176Emperical formula mass of C2H4O = 44gGiven that vapour density = 2×44 = 88gMolecular mass = 2(vapour density)Therefore molecular mass = 2×88 = 176g

The empirical formula of a compound is $C_{2} H_{4} O$ . If the empirical formula mass is equal to one half of its vapor density, find the gram molecular mass of the compound.

The correct option is B 176
Emperical formula mass of $C_{2} H_{4} O$ = 44g
Given that vapour density = $2 \times 44$ = 88g
Molecular mass = 2(vapour density)
Therefore molecular mass = $2 \times 88$ = 176g