Empirical and Molecular Formulas p.16 Complex!!!
1) The molecular formulas of some substances are as follows. Write their empirical
formulas
a) Acetylene, C2H2 (used in oxyacetylene torches). (CH)
b) Glucose, C6H12O6 (the chief sugar in blood). (CH2O)
c) Octane, C8H18 (a component of gasoline). (C4H9)
d) Ammonium nitrate, NH4NO3 (a fertilizer component).(NH4NO3)
2) A radioactive form of sodium pertechnetate is used as a brain-scanning agent
in medical diagnosis. An analysis of a 0.9872 g sample found 0.1220 g of
sodium and 0.5255 g of technetium. The remainder was oxygen. Calculate
the empirical formula of sodium pertechnetate. (Use the value of 98.907 as
the atomic mass of Tc and arrange the atomic symbols in the formula in the
order NaTcO.)
sample xg(Na)
xg(Tc)
xg(O)
(i) Mass(O) = 0.9872 –(0.1220 + 0.5255) = 0.3397 g
Element Mass(g)
Mole
Ratio
Na
0.1220  23.0
= 0.005304348
Tc
0.5255  98
= 0.005362245
1.010 1
O
0.3397 16.0
= 0.02123125
4.002  4
1
Empirical Formula
NaTcO4
3) Potassium persulfate (Anthion®) is used in photography to remove the last traces
of typo from photographic papers and plates. A 0.8162 g sample was found to
contain 0.2361 g of potassium and 0.1936 g of sulfur; the rest was oxygen. The
molecular mass of this compound was measured to be 270. What are the
empirical and molecular formulas of potassium persulfate? (Arrange the atomic
symbols in the formulas in the order KSO.)
sample
xg(K)
xg(S) xg(O)
Mass(O) =0.8162 –(0.2361 + 0.1936)=0.3865
Element Mass(g)
Mole
Ratio
Empirical Formula
K
0.2361  39.1= 0.006038363
S
0.1936  32.1= 0.006031153
1.001=1
KSO4
1
0.3865 16.0 = 0.02415625
4.005=4
MF = (EF) x
MM = (EF) x
270 = (KSO4) x
270 = (39.1 + 32.1 + 64.0) x
270 = 135.2 x
x = 1.997 = 2
Molecular Formula: (KSO4) X 2 = K2S2O8
4) Adenosine triphosphate (ATP) is an important substance in all living cells. A
sample with a mass of 0.8138 g was analyzed and found to contain
O
0.1927 g of carbon, 0.02590 g of hydrogen, 0.1124 g of nitrogen, and 0.1491 g of
phosphorus. The remainder was oxygen. Its molecular mass was determined to be
507. Calculate the empirical and molecular formulas of adenosine triphosphate.
(Arrange the atomic symbols in alphabetical order in the formulas.)
sample xg(C)
xg(H)
xg(N)
xg(P)
xg(O)
Mass(O) = 0.8138 –(0.1927 + 0.02590 + 0.1124 + 0.1491)= 0.3337
Element Mass(g)
Mole
Ratio
Empirical Formula
C
0.1927  12.0 = 0.016058333 3.338 X 3 = 10
H
0.02590  1.0 = 0.02590
5.3849 X 3 = 16
N
0.1124  14.0 = 0.008028571 1.669 X 3 = 5 C10H16N5P3O13
P
0.1491  31.0 = 0.004809677
1 X3 = 3
O
0.3337  16.0 =0.02085625
4.336 X 3 = 13
MM = (EF) x
507 =(C10H16N5P3O13) x
507 = [10(12.0) + 16(1.0) + 5(14.0) + 3(31.0) + 13(16.0)]x
507 = 507x
x=1
EF = MF = Molecular Formula: C10H16N5P3O13
5) Realgar (re-AL-gar) is a deep red pigment used in painting. A 0.6817 g sample
was found to contain 0.4774 g of arsenic; the remainder was sulfur. The
molecular mass of realgar was found to be 428. What are the empirical and
molecular formulas of this pigment? (Arrange the symbols in the order AsS.)
sample xg(As) xg(S)
Mass(S) = 0.6817 –0.4774= 0.2043
Element Mass(g)
Mole
Ratio Empirical Formula
As
0.4774  74.9 = 0.006373832
S
0.2043  32.1= 0.006364486
MF = (EF) x
MM = (AsS) x
428 = (74.9 + 32.1) x
428 = (107) x
x=4
Molecular Formula: (AsS) X 4 = As4S4
1.001=1 AsS
1
6) Isobutylene is a raw material for making synthetic rubber. A sample with a mass of
0.6481 g was found to contain 0.5555 g of carbon; the rest was hydrogen. Its formula
mass was determined to be 57. What are the empirical and molecular formulas of
isobutylene? (Place the atomic symbols in the formulas in the order CH.)
sample xg(C) xg(H)
Mass(H): 0.6481 -0.5555= 0.0926
Element Mass(g)
Mole
C
0.5555  12.0= 0.046291667
H
0.0926  1 =
0.0926
MF = (EF) x
Ratio Empirical Formula
1
CH2
2
MM = (CH2) x
57 = (CH2) x
57 = (12 + 2) x
57 = 14 x
x=4
Molecular Formula: (CH2) X 4 = C4H8 = isobutylene
7) Cyanuric acid is used for such different purposes as making synthetic sponges
and killing weeds. A sample with a mass of 0.5627 g was found to contain 0.1570
g of carbon, 0.01317 g of hydrogen, and 0.1832 g of nitrogen, with the balance
being oxygen. Its molecular mass was found to be 129. Calculate the empirical
and molecular formulas of cyanuric acid, arranging the atomic symbols in
alphabetical order.
Sample xg(C)
xg(H)
xg(N)
xg(O)
Mass(O) = 0.5627 –(0.1570 + 0.01317 + 0.1832) =0.20933 g
Element Mass(g)
Mole
Ratio Empirical Formula
C
0.1570  12.0 = 0.01308333 1.00001 CHNO
H
0.01317  1 = 0.01317
1.00664
N
0.1832  14.0 = 0.013085714 1.0002
O
0.20933  16.0 = 0.013083125 1
MF = (EF) x
MM = (EF) x
129 = (CHNO) x
129 = (1 + 12 + 14 + 16) x
129 = 43 x
x=3
Molecular Formula: (CHNO) X 3 = C3H3N3O3
8) Hypophosphoric acid is one of several acids containing both oxygen and
phosphorus. Its formula mass is 162. A sample with a mass of 0.8821 g was
found to contain 0.0220 g of hydrogen and 0.3374 g of phosphorus, with the
remainder being oxygen. Calculate its empirical and molecular formulas.
(Arrange the atomic symbols in the formulas in the order HPO.)
Mass(O) = 0.8821 – (0.0220 + 0.3374) = 0.5227
Element Mass(g)
Mole
H
0.0220  1
= 0.0220
P
0.3374  31.0 = 0.010883871
O
0.5227  16 = 0.03266875
MM = (EF) x
162 = (H2PO3) x
162 = (2(1.0) + 31.0 + 3(16.0)) x
Ratio Empirical Formula
2.02 H2PO3
1
3
162 = (81) x
x=2
Molecular Formula: (H2PO3) X 2 = H4P2O6
Answer to Questions p.17
1a)CH
b)CH2O
c)C4H9
d)NH4NO3
2)NaTcO4
3)KSO4,
K 2 S 2 O8
4)C10H16N 5P3O13
5)AsS, As4S4
6)empirical:CH2 molecular:
C4H8
7)empirical: CHNO molecular: C3H3N3O3
8)empirical: H2PO3 molecular: H4P2O6