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Name Date Class

PI!It.' Part 0 ProblemsPERCENT COMPOSITION AND., , , , , .

CHEMICAL FORMULAS Solve the fallowing problems in the space provided. Shaw your work.

14. What is the percent composition of each of the [oil owing'

a. Cr,03 c. HgSSection Revgew

Objectives$ Calculate the percent by mass of all element in a compound• Interpret an empirical formula

• Compare and contrast empirical and molecular formulas

0. Mn,P~OTd. CarNO"h

Vocabulary15. Determine the empirical formula of the compound with the percent

compo.ition of 29.1 % Na. 40.5% S. and 30.4% O.

• percent compositionIII empirical formula

Key Equationmass of element

• % mass of element = mass of compound X 100%

16. How many kilograms of iron can be recovered from 639 kilograms of theore Fe203?

Part A Completion 10.3 PERCENT COMPOSiTION AND CHEMICAL FORMULASUse this completion exercise to check your kllowledge of the terms and yourunderstanding of the concepts introduced in I"~'section. Each blank can becompleted with CL term. short phrase, or number.

1. i..sample of a compound analyzed in a chemistry laboratory consists of 5.34 gof carbon, 0.42 g of hydrogen, and 47.08 g of chlorine. What is the percentcomposition of this compound?

2. Find the percent composition of a compound containing tin and chlorine if18.35 g of the compound contains 5.74 g of tin,

3. If 3.907 g of carbon combines completely with 0.874 g of hydrogen to form acompound, what is the percent composition of this compound?

4. Frorn the formula for culciurn acetate, Ca(C,H30,)" calculate the mass ofcarbon that can be obtained frorn 65.3 g 01" the compound.

5. How many grams of aluminum are in 25.0 g of aluminum oxide (A120~)?

6. How many grams of iron are in 21.6 g of iron(IJl) oxide (FeoOs)?

7. Determine the empirical formula of each of the following compounds from thepercent composition:

a. 7.8% carbon and 92.2% chlorine

b. 10.0% C, 0.30% H, 89.1 % C!

The __..L..of a compound is the percent by mass of each 1. _

element ill a compound. The percent by mass of an element in a 2.

compound is the number ofgrarns of the element per __ 2__ g 3.

of the compound, multiplied by 100%. To calculate the percent by 4.

mass of on element in a known compound. divide the mass of the 5.

element in one mole by the __ 3__ and multiply by 100%. 6.

Aln) _4 __ formula represents the lowest __ 5__ ratio of the

elements in a compound. lt can be calculated from a compound's

percent composition. The __ 6_ formuJa of a compound is either

the same as its empirical formula, or it is some whole-number

multiple of it.

Chapter 70ChemicalQuantities 245

Section Review 10.3Part A CompletionL percent composition

2. 100

3. molar mass

4. empirical

5. whole-number6. molecular

Hi. molar mass FeZ03 =

vf, 55.8 g Fe v 16.0 gO2 rnoiFe x, . / + 3 ~O", ~ ~O

1 IT~e 1 1.y>11 ,

= 112 g Fe + 48 g °= 160 g Fe203

a po 112 g%Fe = 0 - v X 100 = --- X 100

g Fe203 160 g

Part D Questions and Problems, 104 g Cr 1. _ II 0"14. a. X -,-00 - 68.",,% 01

152 g Cr203

48 z O- 0 X 100 = 31.6% °

152 g CrZ03

110 g Mnb. X 100 = 38.7% Mn

284 g Mn2P20,62 ap__ ---"'-0 __ X 100 = 21.8% P

284 g Mn2P207112 z O

b X 100 = 39.4% 0284 g Mn2P :P7201 a Hgc. b X 100 = 86.3% Ha

233 g HgS b

32.1 as~T S X 100 = 13.8% S

233 g hg

= 70.0% Fe

70.0 leg Fe = 447 kg Fe639~ x 100.kg-FezCJ3

Section 10.3

40.1 g Ca ')' ,-d. X 100 = ",'±.J% Ca

264 g Ca(N03)2

28 aNb X 100 = 17.1%N

164 g Ca(N03)2

96 gO X 100 = 58.5% °164 g Ca(NOa}2

_ T,X:: 1.00 mol Na ') , ._15. 29.l2A:\la x _"-"" = 1.~7mal Na

'" 23.0 ~a1.27 mol Na/l.27 = 1 x 2 = 2

...(v 1.00 mol S _ .40.5 ~ /, ')~ A = 1.21 mol S

,J~.O ~1.27 mol S/1.27 = 1 X 2 = 2

gD 1.00 malO _30.4 a U X aA'''') = 1.90 mol °

16.0 g-'-/1.90 mol 0/1.27 = 1.5 x 2 = 3

5.34 a C1. Percent C = b d x 100 = 10.1% C

52.84 g cp0.42 aH

Percent H = 0 d x 100 = 0.79% H52.84 g cp47.08 a CI

Percent Cl =. b d X 100 = 89.1 % Cl52.84 g cp

2. Mass of Cl= total mass of compound - mass of Sn= 18.35 g of compound - 5.74 g Sn= 12.61 g Cl

e 5.74 a SnPercent or Sn = 0 d X 100

18.35 g cp= 31.3% Sn

12.61 a ClPercent of CI = 0 d X 100

18.35 g cp

Empirical formula = NazS203

= 68.7% Cl3.907 a C

3. Percent C = 0 d x 100 = 81.7% C4.781 g cp

0.874 g HPercent H = d x 100 = 18.3% H

4.781 g cp~ 48.0 IT C

4. Percent C = b X 100158.1 g Ca(C2H302)2

= 30.4% CMass C = 30.4% C x 65.3 g = 19.8 g

5. 13.2 g Al

6. 15.11 g Fe

7. a. Get!

b. CHCI3