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Percent Composition, Empirical & Molecular Formulas
Percent composition is the percent by mass of each element found in a compound.
Example: What percent of iron(IlI) hydroxide, Fe(OH). is oxygen?
Step 1: Find the molar mass of the compound.
1 mol Fe = 55.85 g
3 molO = 3x16.00g = 48.00 g
3 molH = 3x1.008g = 3.024
Molar Mass Fe(OH)3 = 106.87 g/mo
Step 2: Find the percentage by dividing the part by the whole and multiplying by 100.
48.00¢0
106.87 g Fe(OH),
Part 1: Finding Percent
1. What percent of magnesium bromide, MgBr2, is magnesium?
IMg +2481 glmel = 24.319 2431949 ip d| 13.20% Mg
2&ve TBUADS lwiol = 169.309 184,g Se
14, 11 9 /mot MgBr, 159.8058". $6.79b% Br
2. What percent of glucose, CsH120e, is carbon? 1Z4.1hg
ba t2.Olgluol = 72.0 4C. 72.06 gC : [2.1244
12H hel gloel = l2d2g°H Teg" 89-99% 0. tat we [U1274H)
0 w 1e.08 Gg lmpe = Ib. 005 O
sole eco 0648
\ R018 lmok Cullic 0, eee sae =| §3,28%S
3. What percent of on I BO: "8g
x100 % = 44.91% oxygen
2 is zinc?
B2n «lS out = 146-289 Zn 14.2 3.g2n 6
2P + BOAT gh = bha4g ie 336.179 stg
BO =1bs00 9/9 = 128.009 O bP 100 =) 104 FP athe
1 en on Beta tal SBT.
mi My et hanes,
“aA afet = 52.92% AR 2b AB a 12.87% Ap
ee 13,00.
[AtzOs hos 0. Wiglor Al
Part 2: Find the percent compositions of all of the elements in the following compounds and
name all compounds:
Bs CuBr Name Copper Ci) Bromide
MM = 223.3 7a lmol
63-5544 x100-= 74.a04br 12 = 154.869 OF
223.313 154.809 Br
%CuU: 28,45 % yer: 11,540 % ea —— x lOO =
6. (NHa)2S Name Ammontum, Sultde -
MM = 6B, 1Sq/mot
%6N: qp,iz wot: 1% 96S: 47.0l0
7 NaSp Name dini-ty en. disul-Gde
MM = 92.14 gino
%N: 30.4 | HS: bGlol
Se ALSO.) name Aluminum Sulfate
MM = 3421S qlmo
at: 150772 B xs: ZBAIZ xo: Bb Mlle
Part 3: Empirical Formulas: Find the empirical formula for the following compounds. See page 2
in Baby Blue and pages 86-89 in Big Blue.
9. CeHs
Me jen)
12. CH¢0:
13. XasYi3
Example Problem: A compound with an empirical formula of COH4and a molar mass of 88
grams per mole. What is the molecular formula of this compound?
First find the empirical formula mass of the compound.
Carbon: — 2 (12.01 g/mol)
Oxygen: 1.(16.00 g/mol)
Hydrogen: 4 (1.01 g/mol
44.01 g/mol
So this compound has a molar mass of 88.02 g/mol and an empirical formula mass of
44.01 g/mol. So the molecular formula must have twice as many of each atom as the
empirical formula.
88 /44=2
Molecular formula is 2 times the empirical formula
2(C,OHa) = C4O2He
