Chapter 11 Test Review

Vocab

Conversions

Molar Mass of Compounds

Percent Composition

Empirical Formulas

100

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

Mole

100

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

9.033 x10²³atoms C

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

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

100

What is the formula for percent composition?

(Part/Whole) x 100

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

The mass of one mole of any pure substance is

Molar Mass

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)

200

Find the Molar Mass of C₆ H₁₂O₆

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

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%

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

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

Grams

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)

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

300

A labratory analysis showed that a 120 g sample of NH₃contained 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%

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

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

Atomic Mass

400

How many moles of Iron are in a sample containing 3.8x10²⁷atoms of Iron?

6310.2 mol Fe

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

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

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%

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

Which Formulas has the simplest mole ratio?

Empirical

500

Calculate the mass of .35 mol of Oxygen atoms

5.6 g O

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

500

Find the Molar Mass of C₂ H₆ O₃

78.08 g/mol

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 %

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?

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