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### Question 1

Write a balanced equation (using smallest integer values) for the following reaction:
Mg3N2 + H2O --> NH3 + Mg(OH)2
What is the number of water molecules in the balanced equation?

Select one:

### Explanation

Mg3N2 + H2O --> NH3 + Mg(OH)2

3 Mg

2 N

2 H

1 O

1 Mg

1 N

5 H

2 O

Mg3N2 + H2O --> 2 NH3 + 3 Mg(OH)2

MAKE AN ATTEMPT TO BALANCE

Mg3N2 + 6 H2O --> 2 NH3 + 3 Mg(OH)2

3 Mg

2 N

12 H

6 O

### Question 2

Write a balanced equation (using smallest integer values) for the following reaction:
Fe + O2 --> Fe2O3
What is the number of oxygen molecules in the balanced equation?

Select one:

### Explanation

Fe + O2 --> Fe2O3

4 Fe + 3 O2 --> 2 Fe2O

### Question 3

Write a balanced equation (using smallest integer values) for the following reaction:
Zn + H3PO4 --> H2 +Zn3(PO4)2
What is the number of zinc atoms (or moles) in the balanced equation?

Select one:

### Explanation

Zn + H3PO4 --> H2 +Zn3(PO4)2

3 Zn + 2 H3PO4 --> 3 H2 +Zn3(PO4)2

### Question 4

What is the oxidation state of oxygen in KO2?

Select one:

K= +1

O2= -1

O= -1/2

### Question 5

What is the oxidation state of xenon in XeO2F2?

Select one:

Xe

O2=2(2-)=4-

F2=2(1-)=2-

-6+X=0

X=6

### Question 6

Calculate the mass of carbon in 4.98 g of C4H10O.

### Explanation

MM of C4H10O==(74.12 g/mol)

4.98/74.12=.06718834 moles

4 moles of carbon in C4H10O

.06718834*4=.26875=Xg/ 12 g/mol = 3.2

### Question 7

A 2.770 g sample of a compound containing only carbon, hydrogen, and oxygen burns in excess oxygen to produce 4.06 g of CO2 and 1.66 g of H2O. Calculate the mass (in g) of C in the sample.

### Explanation

3 Sig. Fig.

2.770 g sample C, H, O

4.06 g CO2 / 44.01 g/mol CO2 = 0.092252 mol CO2

Find mass %

C:12  --> 12/44.01= 27.3%

O:16 (2)  --->32/44.01 = 72.7%

0.092252 mol C  = X / (12 g/mol C)

X= 1.11 g C

0.184504 mol O ( there are 2 mols of O in CO2 so we multiply it by 2)

1.66 g H2O/ (18.02 g/mol H2O) = 0.09211987

0.18423973 mol H (2 mols of H in H2O)

.09211987 mol O

0.18423973 mol H  = X g H/ (1.008 g/mol H)

X= 0.18571365

2.770 g sample - 1.11 g C - 0.18571365 g H =1.474 g O

mass % H = 0.18571365 g H / 2.770 g sample * 100% = 6.70%

### Question 8

Calculate the mass (in g) of H in the sample.

1.66 g H2O/ (18.02 g/mol H2O) = 0.09211987 mol H2O

0.18423973 mol H (2 mols of H in H2O)

.09211987 mol O

0.18423973 mol H  = X g H/ (1.008 g/mol H)

X= 0.18571365

### Question 9

Calculate the mass (in g) of O in the sample.

2.770 g sample - 1.11 g C - 0.18571365 g H =1.474 g O

### Question 10

What is the mass percent H in the above sample.

mass % H = 0.18571365 g H / 2.770 g sample * 100% = 6.70%

### Question 11

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of a 2.18 g sample burns in excess oxygen yields 3.94 g of CO2 and 1.89 g of H2O. A separate experiment shows that a 1.23 g sample contains 0.235 g of N. Calculate the moles of C in the sample.

### Explanation

2.18 g C, H, N, O

3.94g CO2/ (44.01 g/ mol CO2) = 0.089525 mol CO2

0.089525 mol C

1.89g H2O/ (18g/mol H2O) = 0.105 mol H20

2(0.105)=0.21 mol H

1.23 g sample contains 0.235 g N

1.23 g sample / 0.235 g N = 2.18 g sample/ X gN

0.4165 = g N

0.4165 g N / (14 g/ mol N) = X mol

0.02975 mol N

0.089525 mol C (12 g/ mol C) = 1.0743 g C

0.21 mol H (1.008 g/ mol H) = 0.21168 g H

0.02975 mol N (14 g/mol N) = 0.4165 g N

1.0743 g C + 0.21168 g H + 0.4165 g N = 1.70248 g C, H, N

2.18 g - 1.70248 g = 0.47752 g O * 1 g O/ (16 g/ mol O)= 0.02985 mol O

Empirical Formula

Divide by the Smallest Mol (in this case N)

0.089525 mol C /0.02975 =3

0.21 mol H / 0.02975 =7

0.02975 mol N / 0.02975 =1

0.02985 mol O/ 0.02975 = 1

Find their rounded/approximate integers

C3H7NO is the final empirical formula

### Question 12

Calculate the moles of H in the sample.

1.89g H2O/ (18g/mol H2O) = 0.105 mol H20

2(0.105)=0.21 mol H

### Question 13

Calculate the moles of N for a 2.18 g sample (the sample size for the other analyses).

1.23 g sample / 0.235 g N = 2.18 g sample/ X gN

0.4165 = g N

0.4165 g N / (14 g/ mol N) = X mol

0.02975 mol N

### Question 14

Calculate the moles of O in the sample.
Hint given in feed back

Hint, first subtract the mass C, H, and N from the total mass to get the mass of O.

### Question 15

Determine the empirical formula. How many moles of H are in the empirical formula?

E.g., in H2O there are 2 moles of H for each mole of water.

0.089525 mol C /0.02975 =3

0.21 mol H / 0.02975 =7

0.02975 mol N / 0.02975 =1

0.02985 mol O/ 0.02975 = 1

Find their rounded/approximate integers

C3H7NO is the final empirical formula

### Question 16

Calculate the mass (in g) of O2 consumed in the complete combustion of 56.3 g sample of C4H8O. (Hint, write the balanced equation.)

### Explanation

2 C4H8O + 11O2 --> 8CO2 + 8H2)

56.3 g/ (72.11 g/mol) = 0.781mol C4H8O * 11 mol O2/ 2mol C4H8O

=4.2941 mol O2=X g O2/ (32 g/mol O2) = 137.41g O2

### Question 17

Carbon disulfide (CS2) reacts with excess chlorine (Cl2) to produce carbon tetrachloride (CCl4) and disulfur dichloride (S2Cl2). If 61.5 g of CS2 yields 87.4 g of CCl4, what is the percent yield? (Hint, you must first write the balanced equation.)

### Explanation

CS2 + 2Cl2 --> CCl4 + S2Cl2

61.5 g / (76.129 g/mol) = 0.8077 mol CS2

87.4 g/ (153.82 g/mol) = 0.5682 mol CCl4

0.8077 mol CS2 * 1 mol CCl4 / 1 mol CS2 = 0.8077 mol CCl4

0.8077 CCl4 = Xg CCl4/ (153.82 g/mol CCl4)

X=124.24

% yield = actual yield/ theoretical yield * 100%

87.4 g/ 124.24 g * 100% = 70.348%

### Question 18

The reaction of 6.82 g of Cl2 with 2.20 g of P4 produces 3.53 g of PCl5. What is the percent yield?
Hint given in feedback

### Explanation

Hint, this is a limiting reactant problem.

10 Cl2 + P4  ---> 4PCl5

6.82 g/ (70.91 g/mol Cl2) = 0.0962 mol Cl2

2.20 g/ (123.9 g/mol P4) = 0.017756 mol P4

3.53 g/ (208.24 g/mol PCl5)= 0.01695 mol PCl5

.0962 mol Cl2 * 1 mol P4/ 10 mol Cl2 = 0.00962 P4 needed

Cl2 is the limited reagent

0.0962 mol Cl2 * 4 mol PCl5/ 10 mol Cl2 = 0.03838 mol PCl5 (208.24 g/mol)

= 8.013 g

3.53 g PCl5/ 8.013 g PCl5 * 100 = 44.05%

### Question 19

The reaction of equal molar amounts of benzene, C6H6, and chlorine, Cl2, carried out under special conditions, yields a gas and a clear liquid. Analysis of the liquid shows that it contains 64.03% carbon, 4.48% hydrogen, and 31.49% chlorine by mass and that is has a molar mass of 112.5 g/mol. The molecular formula will be determined. First, determine the number of moles of carbon in a 100 g sample.

### Explanation

C6H6 + Cl2 --> Gas + Clear Liquid

100g Sample

64.03 g C/ 12 g/mol C)  = 5.34 mol C

4.48 g H/ (1.008 g/mol H) = 4.44 mol H

31.49 g Cl/ 35.453 g Cl = 0.8882182 mol Cl

Divide the mols by the smallest mol to determine the empirical formula

5.34 mol C/ 0.8882182 mol = 6

4.44 mol H/ 0.8882182 mol=5

0.8882182 mol Cl/ 0.8882182 mol=1

C6H5Cl   == EMPIRICAL FORMULA

### Question 20

Second, determine the number of moles of hydrogen in a 100 g sample.

### Question 21

Third, determine the number of moles of chlorine in a 100 g sample.

### Question 22

Determine the empirical formula for this compound. How many moles of hydrogen atoms are in 1 mole of the empirical formula?

### Question 23

Determine the molecular formula for this compound. How many moles of hydrogen atoms are in one mole of molecules (or equivalently how many hydrogen atoms are in one molecule)?

### Question 24

A chemist received a white crystalline solid to identify. When he heated the solid to 500 oC, it did not melt. The solid dissolved in water to give a solution that conducted electricity. Which of the following statements are true?

Select one or more:

### Question 25

What is wrong with the name "chromium trichloride" for the compound CrCl3? (Check all correct statements.) Hint, chromium has more than one oxidation state.

Select one or more:

### Explanation

Rules for Naming Molecular Compounds:

1. Remove the ending of the second element, and add "ide" just like in ionic compounds.
2. When naming molecular compounds prefixes are used to dictate the number of a given element present in the compound. " mono-" indicates one, "di-" indicates two, "tri-" is three, "tetra-" is four, "penta-" is five, and "hexa-" is six, "hepta-" is seven, "octo-" is eight, "nona-" is nine, and "deca" is ten.
3. If there is only one of the first element, you can drop the prefix. For example, CO is carbon monoxide, not monocarbon monoxide.
4. If there are two vowels in a row that sound the same once the prefix is added (they "conflict"), the extra vowel on the end of the prefix is removed. For example, one oxygen would be monooxide, but instead it's monoxide. The extra o is dropped.

### Question 26

What is the correct formula of barium sulfide?

Select one:

### Question 27

What is the correct name of Fe3(PO4)2?

(Polyatomic ions are given in the hints to the first common and in Table 3.5 on page 91 of Tro.)

Select one:

Name the positively charged cation first, followed by the negatively charged anion.

### ROMAN NUMERALS IN IONIC COMPOUND NAMES

A Roman numeral in parentheses, followed by the name of the element, is used for elements that can form more than one positive ion.

Fe2+ Iron(II)
Fe3+ Iron(III)

### NAMING IONIC COMPOUNDS USING   -OUS AND -IC

Although Roman numerals are used to denote the ionic charge of cations, it is still common to see and use the endings -ous or -ic.

Fe2+ Ferrous
Fe3+ Ferric

Example: FeCl3 is ferric chloride or iron(III) chloride.

### NAMING IONIC COMPOUNDS USING   -IDE

The -ide ending is added to the name of a monoatomic ion of an element.

H- Hydride
F- Fluoride
O2- Oxide

### NAMING IONIC COMPOUNDS USING  -ITE AND -ATE

Some polyatomic anions contain oxygen. These anions are called oxyanions. When an element forms two oxyanions, the one with less oxygen is given a name ending in -ite and the one with more oxygen are given a name that ends in -ate.

NO2- Nitrite
NO3- Nitrate

### NAMING IONIC COMPOUNDS USING HYPO- AND PER-

In the case where there is a series of four oxyanions, the hypo- and per- prefixes are used in conjunction with the -ite and -ate suffixes. The hypo- and per- prefixes indicate less oxygen and more oxygen, respectively.

ClO- Hypochlorite
ClO2- Chlorite
ClO3- Chlorate
ClO4- Perchlorate

### IONIC COMPOUNDS CONTAINING BI- AND DI- HYDROGEN

Polyatomic anions sometimes gain one or more H+ ions to form anions of a lower charge. These ions are named by adding the word hydrogen or dihydrogen in front of the name of the anion. It is still common to see and use the older naming convention in which the prefix bi- is used to indicate the addition of a single hydrogen ion.

HCO3- Hydrogen carbonate or bicarbonate
HSO4- Hydrogen sulfate or bisulfate
H2PO4- Dihydrogen phosphate

### Question 28

What is the correct formula of aluminum nitrate?

Select one:

### Question 29

Write the formula for barium hydrogen phosphate.

Select one:

### Question 30

Write the formula for aluminum carbonate.

Select one:

### Question 31

Write the formula for potassium nitrate.

Select one:

### Question 32

Write the formula for sodium sulfate.

Select one:

### Question 33

Check those substances which conduct electricity when dissolved in water.

Select one or more:

### Explanation

The question is asking for strong electrolytes aka things that completely dissolve in water...you can figure which one does which by looking at the solubility rules...

1. Salts containing Group I elements (Li+, Na+, K+, Cs+, Rb+) are soluble . There are few exceptions to this rule. Salts containing the ammonium ion (NH4+) are also soluble.
2. Salts containing nitrate ion (NO3-) are generally soluble.
3. Salts containing Cl -, Br -, or I - are generally soluble. Important exceptions to this rule are halide salts of Ag+, Pb2+, and (Hg2)2+. Thus, AgCl, PbBr2, and Hg2Cl2 are insoluble.
4. Most silver salts are insoluble. AgNO3 and Ag(C2H3O2) are common soluble salts of silver; virtually all others are insoluble.
5. Most sulfate salts are soluble. Important exceptions to this rule include CaSO4, BaSO4, PbSO4, Ag2SO4 and SrSO4 .
6. Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and Al3+ are insoluble. Thus, Fe(OH)3, Al(OH)3, Co(OH)2 are not soluble.
7. Most sulfides of transition metals are highly insoluble, including CdS, FeS, ZnS, and Ag2S. Arsenic, antimony, bismuth, and lead sulfides are also insoluble.
8. Carbonates are frequently insoluble. Group II carbonates (CaCO3, SrCO3, and BaCO​3) are insoluble, as are FeCO3 and PbCO3.
9. Chromates are frequently insoluble. Examples include PbCrO4 and BaCrO4.
10. Phosphates such as Ca3(PO4)2 and Ag3PO4 are frequently insoluble.
11. Fluorides such as BaF2, MgF2, and PbF2 are frequently insoluble

### Question 34

For which of the following compounds are the name and the formula correct?
Hint given in feedback

Select one or more:

### Explanation

Look above...not directly above a little higher than that...possibly as high as you totally kidding we do not condone drug usage #DARE

Question 35

Calculate the number of moles in 6.71x10E22 NO2 molecules.

Question 36

6.71x10E22 molecules x (1 mol / 6.022E23 molecules) = 0.111

Calculate the mass (in g) of 4.98 moles of H2O.

MM H2O :  18.016 g/mol

4.98 mols * 18.016g / mol = 89.7g

Question 37

Calculate the number of moles in 45.4 g SO3.

MM SO3:  45.4 g /mol

(45.4g)  /  ( 1 mol /  80.063g) = .567 mol

Question 38

Calculate the number of oxygen atoms in 10.0 g SO3.

Remember Scientific notation for moodle!

MM SO3:  45.4 g /mol

10g  x (1 mol/ 80.063g) x (3 mol O / 1 mol SO3)  x (6.022E23 atoms/1 mol) = 2.26E23 atoms