Answer:
C2F2Br2O2
Step-by-step explanation:
To determine the molecular formula of the compound given the percentage composition and molar mass, we need to first find the empirical formula and then use the molar mass to calculate the molecular formula.
The empirical formula gives the simplest whole-number ratio of atoms in a compound. We can assume a 100 g sample of the compound to make the percentages directly equal to the mass in grams.
From the given percentage composition, we can calculate the empirical formula as follows:
Mass of carbon (C) = 9.4 g
Mass of fluorine (F) = 14.9 g
Mass of bromine (Br) = 62.7 g
Mass of oxygen (O) = 6.3 g
We can assume that the sample contains 100 g of the compound, so we can convert these masses to moles by dividing by their respective molar masses:
Moles of C = 9.4 g / 12.011 g/mol = 0.783
Moles of F = 14.9 g / 18.998 g/mol = 0.784
Moles of Br = 62.7 g / 79.904 g/mol = 0.784
Moles of O = 6.3 g / 15.999 g/mol = 0.394
Next, we can divide each of the mole values by the smallest mole value to get the mole ratio:
Mole ratio: C: F: Br: O = 0.783 / 0.394 : 0.784 / 0.394 : 0.784 / 0.394 : 0.394 / 0.394
Simplifying the ratio: C2F2Br2O1
So, the empirical formula of the compound is C2F2Br2O1.
To find the molecular formula, we need to know the molar mass of the empirical formula. The molar mass of the empirical formula is:
Molar mass of empirical formula = 2(12.011 g/mol) + 2(18.998 g/mol) + 2(79.904 g/mol) + 15.999 g/mol = 235.8 g/mol
We can calculate the molecular formula by dividing the given molar mass of the compound (237.8 g/mol) by the molar mass of the empirical formula (235.8 g/mol):
Molecular formula = (237.8 g/mol) / (235.8 g/mol) = 1.008
Since the result is very close to 1, we can round up to the nearest whole number. Therefore, the molecular formula of the compound is C2F2Br2O
ALLEN