Chemistry Question when you know the % and amount, how do you transfer that into empirical&molecular formulas?

So the problem is basically:





78.14% boron


21.86% hydrogen





and it can be between 27 and 28 grams [molar mass]





Now how do I solve for the empirical and molecular formulas??|||To get empirical formula, just pretend the % are grams.



So you have

B H

% by Mass (78.14) 21.86

Molar mass ( 10.8 ) 1

Moles (7.235) 21.86

Simplest Mole

Ratio 1 3



Empirical formula: BH3



Now the Molar Mass of BH3 = 13.8g/mol

Find the average for 27 and 28 (because the real molar mass is between these two numbers)

You get 27.5.

Divide 27.5 by 13.8, you get 2



Multiply the empirical formula BH3 by 2.

You get B2H6 and that's the molecular formula.|||okay, so u basically assume that 1% is one gram of the boron and hydrogen. meaning that you will have 78.14g of boron for every mole and 21.86g of hydrogen for every mole. You then divide their mass by their relative atomic mass so

for Boron it would be 78.14g / 10.811 = 7.228

for hydrogen it would be 21.86/ 1 = 21.86

you then have to take the ratio of these two values which would be 7.228:21.86

if you simplify this ratio you get 1 : 3.02 which can be rounded of to 1: 3

this means that for EVERY atom of boron you have 3 atoms of hydrogen

your empiral formula is the simplest ratio of the atoms in a compound so it would be BH3



to find the molecular formula you divide your molar mass( 27-28) by your empirical mass which is 12 rounded off. that would be 27.5/13.8 = 2

you then take this value (2) and multiply each atom in the empirical formula by it so that it will give you B2H6 as your molecula formula.