Determine the molecular formula of a compund which contains 2.19% H, 12.8% C , 85% Br . 1g of compound occupies 119 ml of volume at stp .
C=12 H=1 Br = 80 )

Question

Determine the molecular formula of a compund which contains 2.19% H, 12.8% C , 85% Br . 1g of compound occupies 119 ml of volume at stp .
C=12 H=1 Br = 80 )

anonymous · Answer

my answer is CHBr2 please verify my answer with atomic mass = 188 gram

jfraser · Answer

the empirical formula is CH2Br:
$$12.8g C * (\frac{1mol}{12gC}) = 1.067mol C$$$$85g Br * (frac{1mol Br}{80g Br}) = 1.062mol Br$$ $$2.19g H * (frac{1mol H}{1.001gH}) = 2.18(7)mol H$$
divive each of those by the smallest, and you get a mole ratio of $$CH_{2}Br$$
the mass of that formula is 93g/mol.

You handled the gas data correctly, since the gas data shows that since 1g occupies 119mL; $$\frac{22.4L}{1mol} = \frac{0.119L}{xmol}$$
1g is 0.005 moles of that compound.The molar mass then has to be 188g/mol. The formula you found is the empirical formula, which is the lowest possible ratio of pieces in the molecule. The molecular formula has to have the same ratio of pieces, but add up to 188g/mol.$$\frac{MW}{EW} = n = \frac{MF}{EF}$$
the ratio of molecular weight to empirical weight is $$\frac{188\frac{g}{mol}}{93\frac{g}{mol}} \approx \frac{2}{1}$$ so the molecular formula is roughly twice as large as the empirical formula, or $$2*(CH_{2}Br) = C_{2}H_{4}Br_{2}$$ which has a molar mass of 186g/mol, which is pretty close to 188g/mol.

anonymous · Answer

simply percentage bt aomic wiight for 3 elements. this is the ratios of atomicities for the elements in the compoung. take the ratios to whole number and put in formula. whayou get is emperical formula. 22400 occupeies 188.2. 188.2 divided by emperical frmula weight gives n. this  n should be multiplied throughout to the atomicities of emperical formula to get molecular formula