What is the molecular formula of a compound if its empirical formula is $C F B r O$ $CFBrO$ and its molar mass is 381.01 g/mol?

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What is the molecular formula of a compound if its empirical formula is $C F B r O$ $CFBrO$ and its molar mass is 381.01 g/mol?

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Answer 1 · 2016-05-24T11:17:03

$(C F B r O)$ $(CFBrO)$ $\times$ $xx$ $3$ $3$ = ??

Explanation:

Molar mass of compound ( $381.01$ $381.01$ $g$ $g$ / $m o l$ $mol$ ) has been provided for you.

Now you need to calculate the molar mass of $C F B r O$ $CFBrO$ .
The molar mass of:
$C = 12.0$ $C = 12.0$ $g$ $g$ / $m o l$ $mol$
$F = 19.0$ $F = 19.0$ $g$ $g$ / $m o l$ $mol$
$B r = 79.9$ $Br = 79.9$ $g$ $g$ / $m o l$ $mol$
$O = 16.0$ $O = 16.0$ $g$ $g$ / $m o l$ $mol$

The molar mass of empirical formula, therefore ( $C F B r O$ $CFBrO$ ) is:
$12.0 + 19.0 + 79.9 + 16.0$ $12.0 + 19.0 + 79.9 + 16.0$ = $126.9$ $126.9$ $g$ $g$ / $m o l$ $mol$ .

In order to get the molecular formula of $C F B r O$ $CFBrO$ ,
divide molar mass of compound by molar mass of empirical formula, like this: $\frac{molar mass of compound}{molar mass of empirical formula}$ $"molar mass of compound"/" molar mass of empirical formula"$ = $whole number$ $"whole number"$

i.e. $\frac{381.01}{126.9}$ $381.01/126.9$ = $3.00$ $3.00$
Rounded to nearest whole number

Now, multiply the whole number by the Empirical Formula.
Like this $(C F B r O)$ $(CFBrO)$ $\times$ $xx$ $3$ $3$ = $C_{3} F_{3} B r_{3} O_{3}$ $C_3F_3Br_3O_3$ .

Therefore your molecular formula is $C_{3} F_{3} B r_{3} O_{3}$ $C_3F_3Br_3O_3$ .

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What is the molecular formula of a compound if its empirical formula is $C F B r O$ $CFBrO$ and its molar mass is 381.01 g/mol?

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What is the molecular formula of a compound if its empirical formula is $C F B r O$ $CFBrO$ and its molar mass is 381.01 g/mol?