Unit2NoteExAns

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Unit 2 - Compounds & Stoichiometry Ch. 2, 3, 4, & 12Answers to Examples in Notes

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1 ) Periodic Table2) Chemical formulas

a) Examples Formulas1. Decide which of the following are expected to be ionic and which are expected to be

molecular: (a) KF, (b) NO, (c) SiCl.

(a) Ionic the K (potassium) is a metal.

(b) Covalent both the N (nitrogen) and O (oxygen) are nonmetals.

(c) Covalent both the Si (silicon) and Cl (chlorine) are nonmetals.

2. What is the formula of magnesium nitride?

This name implies that it is ionic, since the Mg is metallic.

Put the Mg and N ions together by criss-crossing the charges notice how the Mg

gets the 3, while the N gets the 2.

MgN3. What is the formula of calcium phosphate?

This name implies that it is ionic, since the Ca is metallic.

Put the Ca and PO ions together by criss-crossing the charges. Here, the

polyatomic ion, PO, needs parentheses because there is more than one ion

needed.

Ca(PO)

4. Give the formula of carbon disulfide.

This name implies that the compound is covalent, since both elements are nonmetals.

Do NOT use ionic charges!! Instead, match the prefix (di-) for the number (2) to the

element it is attached (S).CS

5. Give the formula of dinitrogen tetrafluoride.

This name implies that the compound is covalent, since both elements are nonmetals.

Do NOT use ionic charges!! Instead, match the prefix (di-) for the number (2) to the

element it is attached (N), and the other prefix (tetra-) for the number (4) to the

element it is attached (F).

NF

6. Calcium chloride hexahydrate is used to melt snow from roads. What is the formula of

this compound?

This name implies that it is an ionic compound with water molecules attached.

First, write the formula for the ionic compound by combining the Ca with the Cl.

Then, use the prefix (hexa-) to determine how many waters should be attached to

the molecule (6).

CaCl6 HO

3 ) Nomenclaturea) Examples Naming

1. Name BaO.


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This is a simple ionic compound because Ba is a metal AND has no other ionic charge

than +2. So, name the element Ba (barium). Then, name the second element O

(oxygen) by dropping the ending and adding ide (oxide)

Barium oxide

2. Name Cr(SO).

This is also an ionic compound. However, the transition metal Cr has two ionic

charges, +2 or +3. Therefore, calculate the charge of the Cr atom, knowing the

charge of the SO ion is 2. and the overall charge of any compound is zero.

2(Cr) + 3(-2) = 0

2Cr = +6

Cr = +3

So, the Cr atom must carry the +3 charge.

According to the stock system, the compound could be named chromium(III) sulfate.

Or, it could be named by the traditional system as chromic sulfate.3. Name OF.

This is a binary molecular compound. Name the first element with no prefix. Then,

name the second element with its numerical prefix and an ide ending.

Oxygen difluoride

4. Name SN.

This is a binary molecular compound. Name the first element with a numberical

prefix. Then, name the second element with its numerical prefix and an ide

ending.

Tetrasulfur tetranitride

5. Bromine has an oxyacid HBrO, whose name is bromous acid. What is the name andformula of the corresponding anion?

Bromite, BrO

6. Name HIO.

This is an acid (notice the H at the beginning of the compound). First, determine the

name of the anion (IO = hypoiodite). Now, drop the ite ending and replace it

with ous.

Hypoiodous acid

7. Name HPO

This is an acid (notice the H at the beginning of the compound). First, determine the

name of the anion (PO = phosphate). Now, drop the ate ending and replace it

with ic. Finally, give it a little bit of a slur by throwing a or in there.

Phosphoric acid

8. A compound whose common name is green vitriol has the chemical formula FeSO 7 HO.

What is the chemical name of this compound?

This is an ionic hydrate. However, the transition metal Fe has two ionic charges, +2

or +3. Therefore, calculate the charge of the Fe atom, knowing the charge of the

SO ion is 2. and the overall charge of any compound is zero.

1(Fe) + 1(-2) = 0


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Fe = +2

So, the Fe atom must carry the +2 charge.

According to the stock system, the compound could be named iron(II) sulfate

heptahydrate.

Or, it could be named by the traditional system as ferrous sulfate heptahydrate.

4 ) Balancing Equationsa) Examples Balancing

Balance the following equations:

1. NH + O NO + HO

First, choose the molecule that is the largest (I picked NH) and place a 1 in front of it.

1 NH + O NO + HO

Notice that this now has one N on the reactants AND products sides. However, there are

3 H atoms on the reactants and 2 H atoms on the products. There is no coefficient(even a sensible fraction) that can be placed in front of the HO to make it have 3

atoms. So, double the only coefficient present.

(2x1) NH + O NO + HO

This change affects both the N atoms and the H atoms. Place a 2 in front of the NO

molecule to make 2 N atoms.

2 NH + O 2 NO + HO

What coefficient should be placed in front of the HO? Did you say 3?

2 NH + O 2 NO + 3 HO

Now count up the O atoms on the products side. Did you get 5?

At this point, a fraction can be used in front of the O molecule (i.e., fi/ O), or all of thecoefficients that have been placed can be doubled so that the O molecule receives a

whole number coefficient.

2 NH + fi/ O 2 NO + 3 HO

OR

4 NH + 5 O 4 NO + 6 HO

2. CHOH + O CO + HO

First, choose the molecule that is the largest (I picked CHOH) and place a 1 in front

of it.

1 CHOH + O CO + HO

Notice that there are now three atoms that have changed C, H, and O. Let us ignore

the O atoms for now, since every molecule has oxygen in it.

There are currently 2 C atoms on the reactants side, so place a 2 in front of the CO.

1 CHOH + O 2 CO + HO

The 1 in front of the CHOH also makes 6 H atoms on the reactants side, so place a

3 in front of the HO.

1 CHOH + O 2 CO + 3 HO


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Now, count up the O atoms on the products side 7. Notice that there is 1 O atom on

the reactants side that must be accounted for. This means that 6 more O atoms are

needed on the reactants side. So place a 3 in front of the O molecule.

1 CHOH + 3 O 2 CO + 3 HO

5 ) General Reactionsa) Examples Types of Reactions

Single Replacement (Displacement) Reactions:

1. 2 Li + 2 HOH 2 LiOH + 1 H

2. Ca + HSO CaSO + H

3. 2 Al + 3 Fe(NO) 3 Fe + 2 Al(NO)

4. F + 2 NaI 2 NaF + I

Double Replacement (Displacement) Reactions:

5. 2 NaCl + HSO 2 HCl + NaSO6. NaOH + HCl HO + NaCl

7. CuCl + NaS 2 NaCl + CuS

8. 1 Al(SO) + 3 Ca(OH) 2 Al(OH) + 3 CaSO

Synthesis (Combination) and Decomposition

9. 1 P + 10 O 2 PO (any real phosphorus and oxygen compound works)

10. HCO HO + CO

11. 2 Mg + O 2 MgO

12. 2 Na + F 2 NaF

13. 2 HgO 2 Hg + O

Combustion14. 1 CHOH + 3 O 2 CO + 3 HO

15. 1 CH + 8 O 5 CO + 6 HO

Acid-Carbonate and Acid-Sulfite

16. 1 NaCO + 2 HCl 1 HO + 1 CO + 2 NaCl

17. 1 SrSO + 2 HClO 1 HO + 1 SO + 1 Sr(ClO)

6) Atomic Theory (2.1-2.3)7 ) Atomic Weights (2.4)

a) Examples - Moles1. Calculate the formula weight of methylamine, CHNH, to two decimal places.

12.01 + 3(1.01) + 14.01 + 2(1.10) = 31.07 g/mol

2. What is the mass of the nitric acid molecule, HNO?

1.01 + 14.01 + 3(16.00) = 63.02 amu

3. A sample of nitric acid contains 0.253 mol HNO. How many grams is this?

0.253mol63.02g

1mol

=15.9g


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4. The daily requirement of chromium in the human diet is 1.0x10fl g. How many moles of

chromium does this represent?

1.0x10-6gCr1mol

52.00g

=1.9x10-8molCr

5. The average daily requirement of the essential amino acid leucine, CHON, is 2.2 g for

an adult. How many atoms of leucine are required daily?

2.2gC6H

14O

2N

1mol

132.21g

6.02x10

23molecules

1mol

23atoms

1molecule

= 2.3x10

23atoms

8 ) Stoichiometry (4.6-4.10)a) examples Simple Stoichiometry

(1) Propane, CH, gas burns according to the equationCH(g) + 5 O(g) 3 CO(g) + 4 HO(g)

How many grams of CO are produced when 20.0 g of propane are burned?

Notice that the equation is already balancedJ

convert

to moles

use the

balanced

equation

to change

substances

convert to

the new

substances

mass

20.0gC3H

8

1molC3H

8

44.11gC3H

8

1CO2

3C3H

8

44.01gCO

2

1molCO2

= 59.9gCO2

(2) Phosphine gas reacts with oxygen according to the following equation:

4 PH(g) + 8 O(g) PO(s) + 6 HO(g)

(a) What is the mass of tetraphosphorus decaoxide produced from 12.43 mol of phophiine?

(b) How many liters of PH gas, at STP, would be required to form 7.39x10 molecules of

water?

(c) How many atoms of oxygen gas are required to produce 1.0000 g water.

Notice that the equation is already balancedJ

Use the same three basic steps as outlined above 1) convert to moles, 2) use the

equations mole ratio, and 3) convert to the unknown substances necessary unit.

(a)

12.43molPH3

1P4O

10

4PH3

283.88gP

4O

10

1molP4O

10

= 882.2gP4O10

(b)

7.39x1023moleculesH

2O

1molH2O

6.02x1023moleculesH

2O

4PH

3

6H2O

22.4LPH

3

1molPH3

=18.3LPH3

(c)

1.0000gH2O

1molH2O

18.02gH2O

8O2

6H2O

6.02x1023moleculesO

2

1molO2

2atomsO

1moleculeO2

= 2.3x10

23atomsO


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b) Examples Advanced Stoichiometry(1) In an experiment, 10.0 g NH and 10.0 g CO were reacted according to the equation

2 NH + CO CHNO + HO

What is the maximum quantity (in grams) of urea CHNO, that can be obtained? How

many grams of the excess reactant are left over at the end of the reaction?

answer:

Notice that the equation is already balanced:)

Additionally, there is information about two reactants, not just one of the species.

This means that the limiting reactant must be determined first.

One of the easiest ways of determining the limiting reactant is to use each of the

reactant amounts to solve the moles of one of the products. As long as the same

product is used in each calculation, the answers can be compared. Whichever will

yield the lower amount of product is the limiting reactant.

10.0gNH3

1molNH3

17.04gNH3

1CH

4N

2O

2NH3

= 0.293molCH4N2O

10.0gCO2

1molCO2

44.01gCO2

1CH

4N

2O

1CO2

= 0.227molCH4N2O

The 0.227 moles of urea is the lower amount; therefore, the CO is the limiting

reactant.

0.227molCH4N

2O

60.07gCH4N

2O

1molCH4N

2O

=13.6gCH4N2O

So, 13.6 g of urea is the maximum quantity that can be obtained.

To determine how much of the excess reactant (NH) will be left over, the amountNH used must be calculated. This amount can be found by beginning with the

original amount of the limiting reactant (CO), which was entirely used up in the

reaction.

10.0gCO2

1molCO2

44.01gCO2

1NH

3

1CO2

17.04gNH

3

1molNH3

= 7.74gNH3

This shows that 7.74 g of NH were used up in the reaction from the original 10.0 g

of NH provided.

Simple subtraction will give the amount of NH is left over.

10.0 g 7.74 g = 2.3 g NH left over

(2) Ethylene, CH, reacts with water to form ethyl alcohol, CHO. When 20.0 g of each

reagent is used, 20.0 g of the alcohol is produced. What is the percentage yield for the

reaction?

answer:

First, a balanced equation is needed.

1 CH + 1 HO 1 CHO

Again, there is information about two reactants, so determine the limiting reactant

first.


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20.0gC2H

4

1molC2H

4

28.06gC2H

4

1C

2H

6O

1C2H

4

= 0.713molC2H6O

20.0gH2O

1molH2O

18.02gH2O

1C

2H

6O

1H2O

=1.11molC2H6O

The CH is the limiting reactant, so determine the mass of alcohol produced from

0.713 mol.

0.713molC2H

6O

46.08g

1mol

= 32.9gC

2H

6O

This 32.9 g of alcohol is the theoretical yield, while the 20.0 g given in the problem is

the experimental amount.

20.0g experimental

32.9gtheoretical100 = 60.8%yield

9 ) Stoichiometry in the Laboratory (4.11-4.12)a) Example Gravimetric analysis

(1)A soluble silver compound was analyzed for the percentage of silver by adding sodiumchloride solution to precipitate the silver ion as silver chloride. If 1.583 g of silver

compound gave 1.788 g of silver chloride, what is the mass percentage of silver in the

compound?

Answer:

Nothiing can be done with the 1.583 g of silver compound because the formula is not

known. However, the formula of silver chloride is known, so the calculation can

begin with the 1.788 g of AgCl.

1.788gAgCl1mol

143.27g

= 0.01248molAgCl

It is important to notice that all of the silver ions (that were in the original silver

compound) were transferred to the silver chloride (AgCl). So the Ag ions from the

silver compound can be solved from the AgCl.

0.01248molAgCl1Ag+

1AgCl

= 0.01248molAg

+

The molar mass of Ag ions is the same as the molar mass of elemental Ag, since the

only difference is that one electron has been removed... hardly any mass at all!

0.01248molAg+ 107.87gAg

+

1molAg+

=1.346gAg

+

Now, the mass of Ag ions can be compared to the mass of the silver compound.

1.346gAg+

1.583gcompound

100 = 85.04%Ag

+

b) Example Volumetric analysis


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(1)A dilute solution of hydrogen peroxide is sold in drug stores as a mild antiseptic. Atypical solution was analyzed for the percentage of hydrogen peroxide by titrating it with

potassium permanganate

5 HO + 2 KMnO + 6 H 8 HO + 5 O + 2 K + 2 Mn

What is the mass percentage of HO in a solution if 57.5 g of solution required 38.9 mL

of 0.534 M KMnO?

Answer:

Nothiing can be done with the 57.5 g of solution because it is a mixture of HO and

water. However, the 38.9 mL of 0.534 M KMnO is enough to solve for the moles

of KMnO.

(0.0389 L)(0.534 M) = 0.0208mol KMnO

Now, the moles of KMnO can be used to determine the mass of HO.

0.0208molKMnO45H

2O

2

2KMnO4

34.02gH2O

2

1molH2O

2

=1.77gH2O2

Now, the mass of HO can be compared to the mass of the solution.

1.77gH2O

2

57.5gsolution

100 = 3.07%H2O2

10)Formula Calculations (4.3-4.5)a) example Percent Composition

(1) How many grams of carbon are there in 61.8 g of sucrose, CHO?

Answer:

First, determine the molar mass of sucrose.

12(12.01) + 22(1.01) + 11(16.00) = 322.34 g/mol

Now, take the mass of each of the elements and divide by this molar mass:

C:12(12.01g)

322.34g

100= 44.71%C

H:22(1.01g)

322.34g

100= 6.89%H

O:11(16.00g)

322.34g

100 = 54.60%O

b) examples Empirical formulas(1) The mass percentages of the elements of sodium pyrophosphate are 28.1% Na, 37.8% P,

and 34.1% O. What is the empirical formula of sodium pyrophosphate?

Answer:

First, assume that the percentages are actually grams. Then, solve each element for

its moles (and follow the significant figures).


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Na:28.1g1mol

22.99g

=1.22molNa

P:37.8g1mol

30.97g

=1.22molP

O:34.1g1mol

16.00g

= 2.13molO

Now, divide each of these moles by whichever is the smallest amount (1.22 moles):

1.22molNa

1.22mol:1.22molP

1.22mol:2.13molO

1.22mol

1.00Na:1.00P:1.75O

Notice that significant figures are still being followed. The 1.75 O should not be

rounded up to 2 because it appears to be the exact fraction 1/. Therefore, it

needs to be multiplied by 4 to make it a whole number; so all of these numbers

need to be multiplied by 4.

4(1.00Na:1.00P:1.75O)

4Na:4P:7OSo, the empirical formula is NaPO.

(2) Elemental analysis of 14.00 g of hexamethylene gives 8.69 g C, 1.93 g H, and the rest N.

What is the empirical formula?

Answer:

First, solve each element for its moles (and follow the significant figures).

C:8.69g

1mol

12.01g

= 0.724molC

H:1.93g1mol

1.01g

=1.91molH

N:14.00gCxHyNz-8.69gC-1.93gH= 3.38gN1mol

14.01g

= 0.242molN

Now, divide each of these moles by whichever is the smallest amount (1.22 moles):

0.724molC

0.242mol:1.91molH

0.242mol:0.242molN

0.242mol

3.00C:7.89H:1.00N

Notice that significant figures are still being followed. The 7.89 H can be roundedbecause it is not a simple fraction.

So, the empirical formula is CHN.

c) Example Molecular formula(1) Knowing the empirical formula of hexamethylene (CHN), what is its molecular formula if

its molecular weight is 116 amu?

Answer:


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EmpiricalMass3(12.01)+8(1.01)+1(14.01)

MolecularMass116amu

58amu= 2

The subscripts in the empirical formula should then be multiplied by 2

C3x2H

8x2N

1x2

So, the molecular formula is CHN.

d) Example - Combustion(1) A sample of benzene weighing 342 mg is burned in oxygen and forms 1156 mg of carbon

dioxide. What is the empirical formula of benzene?

Answer:

Beginning with what is known, that the sample is being burned in oxygen, a sample

reaction can be set up.

CxHy + O HO + CO

This reaction isnt very helpful, since there is no way to balance it. Additionally, the

342 mg of benzene cant be used, since the formula for benzene isnt known. So,

begin with the 1156 mg of CO...

1156mgCO2

1molCO2

44.01gCO2

1C

1CO2

= 26.27mmolC

12.01g

1mol

=315.5mgC

All of the C atoms that ended up in the CO came from the CxHy. So, the mass of

CxHy and the mass of C atoms can be used to determine the mass of the H

atoms.

342 mg CxHy 315.5 mg C = 26.5 mg H (really there are only 2 S.F.)

26.5mgH1mol

1.01g

= 26mmolH

Now, the moles of C and H can be compared.

26.27mmolC

26mmol:26mmolH

26mmol

1.010C:1.0H

So, the empirical formula of benzene is CH.

P.S. Yeah, mmol = millimoles. Hey, mole is a SI unit!

11)Concentration Unitsa) Examples Concentration Units

Molality(1) Iodine dissolves in various organic solvents, such as methylene dichloride, in which it

forms an orange solution. What is the molality of I in a solution of 5.00 g iodine, I, in

30.0 g of methylene chloride, CHCl?

answer:


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5.00gI2

1mol

253.81g

= 0.0197molI2

molality = molSolutekgSolvent

= 0.0197molI20.0300kgCH

2Cl

2

= 0.657mI2

Mass Percent

(2)An experiment calls for 336.0 g of a 5.00% aqueous solution of potassium bromide.Describe how you would make up such a solution.

answer:

Notice that this problem uses the word aqueous. This implies that the solution has

the solvent of water, so the KBr must be the solute.

5.00% =xgKBr

336.0gH2O +KBr

100

x =16.8gKBr

To make this solution, 16.8 g of KBr should be mixed with 319.2 g of water.

Mole Fraction

(3)A solution of iodine in methylene chloride, CHCl, contains 1.50 g I and 56.00 g CHCl.What are the mole fractions of each component in the solution?

answer:

1.50gI2

1mol

253.81g

= 0.00591molI2

56.00gCH2Cl

2

1mol

84.93g

= 0.659molCH2Cl2

CCH2Cl2 =0.659molCH

2Cl

2

0.659molCH2Cl

2+ 0.00591molI

2

= 0.00889

b) Examples Converting Between Concentration Units(1) A 3.6 m solution of calcium chloride is used in tractor tires to give them weight; the

addition of CaCl prevents the water from freezing at temperatures above about 20 C.

What are the mole fractions of CaCl and water in such a solution?

Answer:

First, assume that the denominator of the concentration unit provided is 1.

3.6mCaCl2=

3.6molCaCl2

1kgH2ONow, the mass of water can be used to find the moles of water.

1kgH2O =1000gH

2O

1mol

18.02g

= 55.56molH2O

These moles can be used to solve for the mole fractions.

CCaCl2

=3.6molCaCl

2

3.6molCaCl2+ 55.56molH

2O= 0.0608

CH

2O=1- 0.0608 = 0.939


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(2) A solution contains 8.89x10 mole fraction I dissolved in 0.9911 mole fraction of CHCl.

What is the molality of I in the solution?

Answer:

First, assume that the denominator of each of the mole fractions is 1. This makes

the moles of I equal to 8.89x10 and the moles of CHCl equal to 0.9911. Now,

the mass of the solvent, CHCl, can be solved to get the molality.

0.9911molCH2Cl

2

84.93g

1mol

= 84.17gCH2Cl2

molality =molSolute

kgSolvent=

8.89x10-3molI2

0.08417kgCH2Cl

2

=1.06x10-4mI2

(3) Citric acid, HCHO, is often used in fruit beverages to add tartness. An aqueous

solution of citric acid is 2.331 m HCHO. What is the molarity of citric acid in the

solution? The density of the solution is 1.1346 g/mL.Answer:

First, notice that the term aqueous is used, meaning that the solvent is water.

Then, assume that the denominator of the concentration unit provided is 1.

2.331m =2.331molHC

6H

7O

7

1kgH2O

For molarity, the moles of the solute are needed, which is present, AND the volume of

the solution, meaning both the solute and solvent. According to the question, the

only way to obtain any sort of volume is with the density, which is also for the

solution (= solute + solvent). So, the masses of the solute and solvent are needed.

2.331molHC6H7O7192.14

g1mol

=447.9gHC6H7O7

1kgH2O =1000gH

2O

Solution=1000gH2O + 447.9gHC

6H

7O

7=1447.9gSolution

1mL

1.1346g

=1276mLSolution

M=molSolute

LSolution=2.331molHC

6H

7O

7

1.276L=1.827M

(4) An aqueous solution of ethanol, CHOH is 14.1 M. The density of the solution is 0.853

g/cm. What is the molality of ethanol in the solution?

Answer:

First, assume that the denominator of the concentration unit provided is 1.

14.1MC2H

5OH=

14.1molC2H

5OH

1LSolution=

14.1molC2H

5OH

1LC2H

5OH+ H

2O

To solve for the molality, the mass of the solvent is needed. The volume of the

solution can be used to solve for the mass of the solution (= solvent + solute) to

get the mass of the solvent alone.


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14.1molC2H

5OH

46.08g

1mol

= 650.gC2H5OH

1000mLC2H

5OH+H2O 0.8539gmL

= 853.9gC2H5OH+H2O

(853.9gC2H

5OH+ H2O) - 650.gC2H5OH= 203.gH2O

m =molSolute

kgSolvent=14.1molC2H5OH

0.203kgH2O

= 69.5m

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