Resources:

Have you found videos, websites, or explanations that helped you understand this chapter? Let us know and we'll add them to "Resources" part of this page for other students to use.

Question 1

A compound is found to consist of 28.1% sodium, 13.2% boron and 58.7% oxygen. What is its simplest formula?

Select one: Explanation

Na: 28.1g / (23 g/mol) = 1.22 mol

B: 13.2 g/ (10.81 g/mol) = 1.22 mol

O: 58.7g/ ( 16 g/mol) = 3.67 mol

Na:  1.22 / 1.22 = 1

B: 1.22 / 1.22 = 1

O : 3.67 / 1.22 = 3

NaBO3

Question 2

Which of these compounds contains the greatest mass percentage of nitrogen?

Select one: Explanation

CH3NH2 = 14 / (12 +3+14+2) = 0.45

NaNH2 = 14 / (14 +35 +2) = 0.27

Al(CN)3 =  14 *3 / (27 + 36 + 42) = 0.4

Pb(N3)2  = 84 / (207+84) = 0.28

Question 3

What mass of carbon is present in 0.500 mol of phenol (C6H5OH)?

Select one: Explanation

molar mass of phenol = 94.11 g/mol

mass = molar mass x moles = 94.11 x 0.500 = 47.06 g

the mass percentage of C in phenol = 6x12/94.5 = 76.2%

mass of carbon = 47.06 x 76.2% = 36 g

Question 4

A 4.000 g sample of an unknown metal, M, was completely burned in excess O2 to yield 0.02225 mol of the metal oxide, M2O3. What is the metal?

Select one: Explanation

4M + 3O2 ---> 2M2O3

from this reaction we can see that the moles of M = 2 * moles of M2O3
= 2 * 0.02225 = 0.0445 mol

mass of M = 4.000 g

molar mass of M = mass / moles = 4.000 / 0.0445 = 89 g/mol

the molar mass of Y = 89 g/mol

Question 5

What is the maximum mass in grams of aluminum chloride that could be obtained from 2.50 mol of barium chloride and excess aluminum sulfate? This is the balanced equation for the reaction: Al2(SO4)3 + 3BaCl2--> 3BaSO4 + 2AlCl3. BaCl2 molar mass is 208.3 g/mol and AlCl3molar mass is 133.3 g/mol.

Select one: Explanation

Al2(SO4)3 + 3BaCl2--> 3BaSO4 + 2AlCl3

from this equation we know that molar ratio of BaCl2 to AlCl3 = 3:2

2.5 mol BaCl x ( 2 mol AlCl / 3 mol BaCl2)  = 1.67 mol

molar mass of AlCl3 = 133.3 g/mol

mass of AlCl3 = moles x molar mass = 1.67 x 133.3 = 222 g

Question 6

A sample of a compound of xenon and fluorine contains molecules of a single type; XeFn, where n is a whole number. If 5.0 x 1020 of these XeFn molecules have a mass of 0.172 g, what is the value of n?

Select one: Explanation

1 mole = 6.02 x 1023  molecules

moles of XeFn = (5 x1020 ) x (1mol/  (6.02 x 1023 molecules)= 8.3 x 10-4 mol

mass of XeFn = 0.172 g

molar mass of XeFn =  0.172 g / (0.00083 mol)  = 207.2 g/mol

atomic mass of Xe = 131.3; atomic mass of F = 19.0

207.2 g/mol - 131 g/mol = 75.7 g/mol

(75.7 g/mol) / (19g Fe) = 4

Question 7

1.0 x 10-5 g sample of a compound is known to contain 1.36 x 1017 molecules. This compound is

Select one: Explanation

1 mole = 6.02 x 1023

# moles of compound: 1.36 x 1017 / (6.02 x 1023) = 2.26 x 10-7 mols

molar mass compound:  1.0 x 10-5 / (2.26 x 10-7) = 44.2 g/mol

CO2 = 44.2 g/mol

Question 8

1.438 g sample of a compound of nitrogen and oxygen contains 1.000 g of oxygen. What is the empirical formula?

Select one: Explanation

N : 0.438g / 14.01 g/mol = 0.0313  --- > 0.313 /.0313 = 1

O : 1.00g / 16.00 g/mol = .0625 ---> .0625/ .0313 = 2

NO2

Question 9

The limiting reagent in a particular reaction can be recognized because it is the reagent that

Select one: Explanation

The limiting reagent produces the smallest quantity of the product. Therefore, it is the one that is used up first.

Question 10

Consider this reaction used for the production of lead. What is the maximum mass of lead that can be obtained by the reaction of 58.0 g PbO and 30.0 g ofPbS? Select one: Explanation Question 11

What volume of 0.20 M BaCl2 is required to react completely with 25.0 mL of 0.600 M Na2SO4? This is the net ionic equation for the reaction: Select one: Explanation

MBaCl2VBaCl2 = MNa2SO4VNa2SO4

VBaCl2 = MNa2SO4VNa2SO4/MBaCl2 = (25.0 x 0.600) / 0.20 = 75 ml

75 ml

Question 12 Select one: Explanation

PV = nRT

n = PV/ RT

n = (1* 5L) / ( .08206 * 273) = 0.223 mol CaCO3

0.223 mol CaCO3 x ( 1 mol CO2/ 1 mol CaO ) = 0.223 mol

0.223 mol x 100 g/ mol = 22 g