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# Percentage Composition PowerPoint PPT Presentation

part whole. 24 g 95 g. % = x 100. % Mg = x 100. 24.305. 35.453. Mg. Cl. 12. 17. magnesium. chlorine. Percentage Composition. (by mass...not atoms). 25.52% Mg. Mg 2+ Cl 1-. 74.48% Cl. MgCl 2. It is not 33% Mg and 66% Cl. 1 Mg @ 24.305 amu = 24.305 amu

Percentage Composition

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part

whole

24 g

95 g

% = x 100

% Mg = x 100

24.305

35.453

Mg

Cl

12

17

magnesium

chlorine

### Percentage Composition

(by mass...not atoms)

25.52% Mg

Mg2+ Cl1-

74.48% Cl

MgCl2

It is not 33% Mg and 66%Cl

1 Mg @ 24.305 amu = 24.305 amu

2 Cl @ 35.453 amu = 70.906 amu

95.211 amu

### Empirical and Molecular Formulas

A pure compound always consists of the same elements combined in the same proportions by weight.

Therefore, we can express molecular composition as PERCENT BY WEIGHT.

Ethanol, C2H6O

52.13% C

13.15% H

34.72% O

/ 0.708 mol

32.38 g Na

22.65 g S

44.99 g O

/ 0.708 mol

/ 0.708 mol

### Empirical Formula

Quantitative analysis shows that a compound contains 32.38% sodium,

22.65% sulfur, and 44.99% oxygen.

Find the empirical formula of this compound.

sodium sulfate

= 2 Na

32.38% Na

22.65% S

44.99% O

= 1.408 mol Na

Na2SO4

Na2SO4

= 0.708 mol S

= 1 S

= 2.812 mol O

= 4 O

Step 1) %  g

Step 2) g  mol

Step 3) mol

mol

/ 1.19 mol = 1 Na

/ 1.19 mol = 1 H

NaHCO3

/ 1.19 mol = 1 C

/ 1.19 mol = 3 O

### Empirical Formula

A sample weighing 250.0 g is analyzed and found to contain the following:

27.38% sodium

1.19% hydrogen

14.29% carbon

57.14% oxygen

27.38 g Na

1.19 g H

14.29 g C

57.14 g O

Assume sample is 100 g.

Determine the empirical formula of this compound.

Step 1) convert %  gram

Step 2) gram  moles

Step 3) mol / mol

/ 6.917 mol = 1 C

CH2.5

/ 6.917 mol = 2.5 H

(2.4577 H)

### Empirical & Molecular Formula

(contains only hydrogen + carbon)

(~17% hydrogen)

A 175 g hydrocarbon sample is analyzed and found to contain ~83% carbon.

The molar mass of the sample is determined to be 58 g/mol.

Determine the empirical and molecular formula for this sample.

Determine the empirical formula of this compound.

2 C @ 12 g = 24 g

5 H @ 1 g = 5 g

29 g

Step 1) convert %  gram

Step 2) gram  moles

Step 3) mol / mol

Assume sample is 100 g.

Then, 83 g carbon and 17 g hydrogen.

MMempirical = 29 g/mol

C2H5

MMmolecular = 58 g/mol

58/29 = 2

Therefore 2(C2H5) = C4H10

butane

17.8

17.8

0.555 mol

0.555 mol

### Common Mistakes when Calculating Empirical Formula

Given: Compound consists of 36.3 g Zn and 17.8 g S.

Find: empirical formula

36.3 g Zn

= 2 Zn

Zn2S

Chemical formula

indicates MOLE ratio,

not GRAM ratio

17.8 g S

= 1 S

1 mol Zn

1

36.3 g Zn

Zn

= 0.555 mol Zn

65.4 g Zn

ZnS

17.8 g S

1 mol S

1

S

zinc sulfide

= 0.555 mol S

32.1 g S

### Empirical Formula of a Hydrocarbon

1 mol CO2

44.01 g

x

2 mol C

1 mol CO2

x

burn

in O2

g CO2

mol CO2

mol C

mol H

Empirical

formula

CxHy

g H2O

mol H2O

2 mol H

1 mol H2O

x

1 mol H2O

18.02 g

x

Kotz & Treichel, Chemistry & Chemical Reactivity, 3rd Edition , 1996, page 224

/ 183.4 g x 100% = 35.6 % Zn

/ 183.4 g x 100% = 26.2 % C

/ 183.4 g x 100% = 3.3 % H

/ 183.4 g x 100% = 34.9 % O

Find the molar mass and percentage composition of zinc acetate

CH3COO1-

Zn2+

acetate = CH3COO1-

Zn(CH3COO)2

1 Zn @ 65.4 g/mol = 65.4 g

4 C @ 12 g/mol = 48 g

6 H @ 1 g/mol = 6 g

4 O @ 16 g/mol = 64 g

183.4 g

Zn(CH3COO)2

/ 0.5118 mol

/ 0.5118 mol

A compound is found to be 45.5% Y and 54.5% Cl.

Its molar mass (molecular mass) is 590 g.

Assume a 100 g sample size

a) Find its empirical formula

1 mol Y

45.5 g Y

= 0.5118 mol Y

= 1 Y

88.9 g Y

YCl3

1 mol Cl

54.5 g Cl

= 1.535 mol Cl

= 3 Cl

35.5 g Cl

1 Y @ 88.9 g/mol = 88.9g

b) Find its molecular formula

3 Cl @ 35.5 g/mol = 106.5 g

= 3

590 / 195.4

195.4 g

YCl3

3 (YCl3)

Y3Cl9

part

whole

% = x 100 %

6.02x1023

Molar Mass vs. Atomic Mass

2 g

H2 = _____

H2 = _______

2 amu

18 g

H2O = _____

H2O = ________

18 amu

120 g

MgSO4 = _____

MgSO4 = ________

120 amu

149 g

(NH4)3PO4 = _____

(NH4)3PO4 = ________

149 amu

(by mass)

Percentage Composition

Empirical Formula

• %  g

• g  mol

• mol

• mol

Empirical vs. Molecular Formula

(lowest ratio)