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Transcript
```Finding Empirical and Molecular Formulas
1. Convert all masses to moles of each element. If %-mass are given, assume 100 g of the sample (so that % =
massing) and then convert to number of moles using the respective atomic masses.
2. Find which number of moles is the smallest and divide all the different numbers of moles by that smallest number.
If the ratios you obtain are all whole numbers, use the numbers as subscripts to give the formula. This is the
empirical formula!
3. If you do not get whole numbers, then multiply everything by a suitable number that removes the fraction.
Common fractions
If number ends in:
multiply by:
0.25 (1/4)
x4
0.333 (1/3)
x3
0.5 (1/2)
x2
0.666 (2/3)
x3
0.75 (3/4)
x4
4. To get the molecular formula, you also need the approximate Molar Mass (aprox MM) of the compound (this mass
may be determined experimentally and it should be a piece of information in the problem) and the Molar Mass of the
Empirical Formula (MMEF), which you can easily calculate. Get the ratio of both molar masses as:
n
aproxMM
MMEF
Multiply the Empirical Formula by this number n to get the Molecular Formula.
You must round n to the closest integer.
EXAMPLE:
Q. A compound is 78.14% B and 21.86%H by mass. Its molar mass is approx. 30 g/mol. Determine the empirical and
molecular formulas.
1. Assume 100g sample so that % goes directly to mass in grams and then convert to number of mol.
1molB
78.14 gB
 7.2285molB
10.81gB
1molH
21.86 gH
 21.687molH
1.008gH
2. Divide by smallest #mol (in this case the moles of B)
7.2285molB
21.687molH
3H
and
3
 1B
7.2285molB
1B
7.2285molB
Since we obtained whole numbers for the ratios, these are the numbers to use for the empirical formula.
Empirical Formula: BH3
3. (We skip this part of the procedure since we got whole numbers previously!)
4. aproxMM = 30 g/mol,
MMEF 10.81  3(1.008)  13.834g/mol
aproxMM
30 g / mol
round to 2.

 2.168
MMEF
13.834 g / mol
So, the molecular formula is the empirical formula multiplied by 2:
n
Reyes
Molecular Formula is B2H6.
```