Chapter 11 Test Review

Vocab

100

The amount of any substance that contains exactly 6.022x 1023 particles is a

Mole

200

The mass of one mole of any pure substance is

Molar Mass

300

When doing conversions, such as Mass to Moles, what is the unit for mass?

Grams

400

The total mass of an atom of a chemical element is called

Atomic Mass

500

Which Formulas has the simplest mole ratio?

Empirical

Conversions

100

How many atoms are in a 1.5 mol sample of Carbon?

9.033 x10²³ atoms C

( 1.5 mol C ( 6.022x10²³atoms C / 1 mol C )

200

Calculate the mass of a 2.75 mol sample of Carbon

33.03 grams C

2.75 mol C ( 1 mol C / 12.01 g C) 

300

How many atoms are in 25 grams of Nitrogen?

1.07 x 10²⁴ atoms N

25 g N ( 1 mol N / 14.01 g N ) ( 6.022 x 10²³ atoms N / 1 mol n)

400

How many moles of Iron are in a sample containing 3.8x10²⁷ atoms of Iron?

6310.2 mol Fe

3.8 X 10^27 atoms Fe ( 1 mol Fe / 6.022x10²³ atoms Fe)

500

Calculate the mass of .35 mol of Oxygen atoms

5.6 g O

.35 mol O ( 16.00 g O / 1 mol O )

Molar Mass of Compounds

100

Find the Molar Mass of H₂O

18.02 g/mol

2 mol H ( 1.01 g H / 1 mol H ) = 2.02 g H

1 mol O ( 16.00 g O / 1 mol O) = 16.00 g O

2.02 g H + 16.00 g O = 18.02 g/mol H₂O

200

Find the Molar Mass of C₆ H₁₂ O₆

180.16 g/mol

C = 6 moles ( 12.01 g C / 1 mol C )  = 72.066 g C

H = 12 moles (1.01 g H / 1 mol H) = 12. 12 g H

O = 6 moles (16.00 g O / 1 mol O) = 96 g O

72.066 g + 12.12 g H + 96 g O = 180/186 g/ mol C₆ H₁₂ O₆

300

Find the Molar Mass of Cu₂(Cl O₃)₂

294 g/mol

2 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 127.1 g/mol Cu

2 mol Cl ( 35.45 g Cl / 1 mol Cl) = 70.9 Cl

6 mol O ( 16.00 g O / 1 mol O ) = 96 g O

127.1 + 70.9 + 96 = 294 g/mol

400

Find the Molar Mass of Cu₃(NH₄)2

226.714 g/mol

Cu 3 moles ( 63.54 g Cu / 1 mol Cu ) = 190.62 g CuN 2 moles ( 14.01 N / 1 mol N ) = 28.014 g N

H 8 moles ( 1.01 g H / 1 mol ) = 8.08 g H

190.62 + 28.014 + 8.08 = 226.714 g/ mol

500

Find the Molar Mass of C₂ H₆ O₃

78.08 g/mol

Percent Composition

100

What is the formula for percent composition?

(Part/Whole) x 100

200

A labratory analysis of a 20g sample of H₂O showed that it contained 7.5 g of Hydrogen and 12.5g of Oxygen. What is the pecent composition of this compund?

H = 37.5%

O = 62.5%

H( 7.5 g H / 20 g) x 100 = 37.5%

O (12.5 g O / 20 g) x 100 = 62.5%

300

A labratory analysis showed that a 120 g sample of NH₃ contained 45.2 g of Nitrogen and 74.8 g of Hydrogen. What is percent composition?

Nitrogen = 37.7%

Hydrogen = 62.3%

N ( 45.2 g N / 120 g ) x 100 = 37.7%

O (74.8 g H / 120 g H ) x 100 = 62.3%

400

What is the Percent Composition of Cu₂S

Cu = 80%

S = 20%

2 mol Cu ( 63.55  g Cu ) = 127.1 g 

1 mol S (32.06 g S ) = 32.06 g

127.1 + 32.06 = 159.16 g/mol

Cu ( 127.1 / 159.16 ) x 100 = 80%

S ( 32.06 / 159.16 ) x 100 =  20%

500

What is the percent composition of Pb Cl₂

Pb = 74.5%

Cl = 25.5%

1 mol Pb (207.29 g Pb) = 207.2

2 mol Cl (35.45 g Cl ) = 70.9

207.2 + 70.9 = 278 g/mol

Pb (207.2 / 278 ) x 100 = 74.5%

Cl ( 70.9 / 278 ) x 100 = 25.5 %

Empirical Formulas

100

A labraotry analysis of an uknown gas showed that the gas is 82.55% Nitrogen and 17.45% Oxygen by mass. What is the empirical formula of the compund?

N₅O

82.55 % N -> 82.5 g N

17.45 % O -> 17.45 O

82.55 g N ( 1 mol N / 14.01 g N ) = 5.8922 mol N

17.45 g O (1 mol O / 16.00 g O) = 1.0906 mol O

5.8922 / 1.0906 = 5

1.0906 / 1.0906 = 1

N₅O

200

A labratory analysis of an uknown gas showed that the gas is 64.25% Carbon and 35.75% Oxygen by mass. What is the empirical formula of the compound?

OC₂

64.25% C -> 64.25 g C

35.75 % O -> 35.75 g O

64.25g ( 1 / 12.01 ) = 5.3497

35.75 (1 / 16.00 0 = 2.2344

5.3497 / 2.2344 = 2

2.2344 / 2.2344 = 1

300

A labratory analysis showed that it contained 60% Carbon and 40% Hydrogen by mass. What is the empirical formula?

C H₈

60 % C -> 60 g C

40 % H - > 40 g H

60 g C (1 / 12.01 ) = 4.9958

40 g H (1 / 1.01 ) = 39.6048

4.9958 / 4.9958 = 1

39.6048 / 4.9958 = 8

CH₈

400

A labaratory analysis shwoed that a compund contained 80% Carbon and 20% Hydrogen by mass. What is its empirical formula?

CH₃

80% C -> 80 g C

20 % H - > 20 g H

80g C ( 1/ 12.01) = 6.661 mol C

20 g H (1 / 1.01 g H) = 19.80 mol H

6.661 / 6.661 = 1

19.80 / 6.661 = 3

CH₃

500

A labratory analysis of an uknown substance showed that it contained 63.45% Oxygen and 36.55% Hydrogen by mass. What is the empirical formula?

OH

63.45 % - > 63.45 G O

36.55 % -  > 36.55 g H 

63.45 g O (1 mol O/ 16.00 g O ) = 3.9656 mol O

36.55 g H ( 1 mol H  / 1.01g H) = 36.1881 mol H

3.9656 / 3.9656 = 1

3.9656 / 36. 1881 = 1

OH