NOTES – MOLAR MASS AND MOLE CONVERSIONS



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CALCULATING MOLAR MASS OF COMPOUNDS

a. sodium hydroxide FORMULA: ______ NaOH __________________________

Na = 23.0g

O = 16.0g

H = 1.0g

23.0 + 16.0 + 1.0 = 40.0 g/mol NaOH

b. calcium cyanide FORMULA: ____ Ca(CN)2____________________________

Ca = 40.1g

C = 2 (12.0g) = 24.0g

N = 2 (14.0g) = 28.0g

40.1 + 24.0 + 28.0 = 92.1 g/mol Ca(CN)2

c. magnesium phosphate FORMULA: _____ Mg3(PO4)2___________________________

Mg = 3 (24.3g) = 72.9g

P = 2 (31.0g) = 62.0g

O = 8 (16.0g) = 128.0g

72.9 + 62.0 + 128.0 = 262.9 g/mol Mg3(PO4)2

d. iron (III) chromate FORMULA: ___Fe2(CrO4)3_____________________________

Fe = 2 (55.8g) = 111.6g

Cr = 3 (52.0g) = 104g

O = 12 (16.0g) = 192.0g

111.6 + 104 + 192 = 407.6 g/mol Fe2(CrO4)3

PER CENT COMPOSITION

a. Find the per cent composition of calcium cyanide:

Formula: ____ Ca(CN)2______________________

% Ca = 40.1g/92.1g x 100 = 43.5%

% C = 24.0g/92.1g x 100 = 26.1%

% N = 28.0g/92.1g x 100 = 30.4%

% Ca = __43.5%________

% C = __26.1%________

% N = __30.4%________

How many grams of calcium are in 40.0g of calcium cyanide?

43.54 g Ca = x

100 g Ca(CN)2 40.0 g

x = 17.4 g Ca

b. Find the per cent composition of potassium sulfate:

Formula: _____K2SO4_____________________

% K = 78.2g/174.3g x 100 = 44.9%

% S = 32.1g/174.3g x 100 = 18.4%

% O = 64g/174/.3g x 100 = 36.7%

% K = __44.9%________

% S = __18.4%________

% O = ___36.7%_______

How many grams of potassium are in 350g of potassium sulfate? ___________________

78.2g K = x

100 g K2SO4 350g K2SO4

x = 273.7g K

What mass of potassium sulfate contains 15g of potassium? ___________________

78.2g K = 15g K

100 g K2SO4 x

x = 19.2 g K2SO4

TRY:

1. Find the molar mass of each compound below:

a. lithium carbonate ____________________

FORMULA: Li2CO3

Li = (2)6.9g

C = 12.0g

O = 3 (16.0g) = 48.0g

(2)6.9 + 12.0 + 48.0 = 73.8 g/mol Li2CO3

b. calcium nitrate _____________________

FORMULA: Ca(NO3)2

Ca = 40.1g

N = 2 (14.0g) = 28.0 g

O = 6 (16.0g) = 96.0 g

40.0 + 28.0 + 96.0 = 164.1 g Ca(NO3)2

c. tin (IV) sulfate ______________________

FORMULA: Sn(SO4)2

Sn = 118.7g

S = 2 (32.1 g) = 64.2g

O = 8 (16.0g) = 128.0g

118.7 + 64.2 + 128.0 = 310.9 g Sn(SO4)2

2. Find the percent composition of tin (IV) sulfate.

FORMULA: Sn(SO4)2

% Sn = 118.7g/310.9g x 100 = 38.2%

% S = (2 x 32.1g)/310.9g x 100 = 20.6%

% O = (8 x 16.0g)/310.9g x 100 = 41.2%

% Sn = _38.2%_________

% S = _20.6%_________

% O = _41.2%_________

How many grams of tin are in 250g of tin (IV) sulfate?

38.18 g Sn = x

100 g Sn(SO4)2 250 g

x = 95.5 g Sn(SO4)2

MOLE CONVERSIONS

1. Find the molar mass of each compound below:

a. diphosphorus pentoxide __142 g/mol ____________

P2O5

b. copper (II) sulfate __159.6 g/mol ___________

CuSO4

c. calcium cyanide ___92.1 g/mol __________

Ca(CN)2

d. aluminum nitrate __213 g/mol ___________

Al(NO3)3

MOLE CONVERSIONS:

6.02 x 1023 molecules = 1 mole (used for molecular compounds)

6.02 x 1023 formula units = 1 mole (used for ionic compounds)

molar mass in grams = 1 mole (add up molar mass from periodic table)

1. Find the mass of 4.50moles of diphosphorus pentoxide. _____________

4.5 moles 142 g = 639 g

1 mole

2. How many moles is 250.0g of copper (II) sulfate? _____________

250.0 g 1 mole = 1.57 moles

159.5 g

3. Find the mass of 0.545moles of calcium cyanide. ______________

0.545 mole 92.1 g = 50.2 g

1 mole

4. How many molecules are in 110g of diphosphorus pentoxide? _______________

110 g 1 mole 6.02 x 1023 molecules = 4.7 x 1023 molecules

142 g 1 mole

Try: Make the following mole conversions.

STEP 1: Write the correct formula

STEP 2: Determine the molar mass

STEP 3: Use dimensional analysis to convert

a. 575g of sodium sulfate to moles ________________

Na2SO4 (2)(23) + 32 + (4)(16) = 142 g/mole

575 g 1 mole = 4.05 moles

142 g

b. 0.025moles of phosphorus pentachloride to grams ________________

PCl5 (31) + (5)(35.5) = 208.5 g/mole

.025 moles 208.5 g = 5.21 g

1 mole

c. 15.0g of iron(III) nitrate to moles ________________

Fe(NO3)3 55.8 + (3)(14) + (9)(16) = 241.8 g/mole

15.0 g 1 mole = 0.062 moles

` 241.8 g

d. 8.02 x 1023 molecules of carbon disulfide to grams ________________

CS2 12 + (2)(32.1) = 76.2 g/mole

8.02 x 1023 molecules 1 mole 76.2 g = 101.5g

6.02 x 1023 molecules 1 mole

MOLES AND VOLUME

AVOGADRO’S HYPOTHESIS: equal volumes of gases at the same temperature and pressure contain equal numbers of particles

STP: standard temperature and pressure

0°C (temperature) 1 atm (pressure)

TRY: Make the following mole conversions.

a. 14.0L of nitrogen gas at STP to moles

14.0 L N2 1 mole = 0.625 moles

22.4 L

b. 2.5g of chlorine gas at STP to molecules

2.5 g Cl2 1 mole 6.02 x 1023 molecules = 2.1 x 1022 molecules

71.0 g Cl2 1 mole

c. 2.24 x 1025 atoms of neon at STP to liters

2.24 x 1025 atoms Ne 1 mole 22.4 L = 833 L

6.02 x 1023 atoms 1 mole

d. 13.3L of fluorine gas at STP to grams

13.3 L F2 1 mole 38.0 g F2 = 22.6 g F2

22.4 L 1 mole

EMPIRICAL AND MOLECULAR FORMULAS

1. a. Find the percent composition of C2H4.

%C = __85.7%_____

%H = __14.3%_____

b. Find the percent composition of C5H10.

%C = __85.7%_____

%H = __14.3%_____

empirical formula: a formula with the lowest whole number ratio of elements in a compound - simplified (CH2)

molecular formula: a chemical formula of a molecular compound that shows the kinds and number of atoms present in a molecule of a compound – NOT simplified (C2H4) and (C5H10)

TRY: Determine the empirical formula of each compound below.

a. P4O10 b. C6H12O6 c. C3H6O

P2O5 CH2O C3H6O

1. The empirical formula of a compound is CH2O. Its molar mass is 360g/mol. Find its molecular formula.

CH2O = 12 + 2(1) + 16 = 30 g/mol

360 / 30 = 12 ( C12H24O12

2. The empirical formula of a compound is P2O5. Its molar mass is 284g/mol. Find its molecular formula.

P2O5 = 2(31.0) + 5(16) = 142 g/mol

284 / 142 = 2 ( P4O10

MOLAR MASS AND % COMPOSITION HOMEWORK NAME:

DATE:

1. Label each compound below as ionic or molecular. Write the formula and determine the molar mass of each compound.

|COMPOUND |IONIC/ |FORMULA |MOLAR MASS |

| |MOLECULAR? | | |

| |I |BaSO4 |233.4 g/mol |

|barium sulfate | | | |

| |C |C3H8 |44 g/mol |

|tricarbon octahydride | | | |

| |I |Fe2(CO3)3 |291.6 g/mol |

|iron (III) carbonate | | | |

| |C |P4O10 |284 g/mol |

|tetraphosphorus decoxide | | | |

| |I |Sr3(PO4)2 |452.8 g/mol |

|strontium phosphate | | | |

2. Find the per cent composition of iron (III) carbonate.

% Fe = 38.3%

% C = 12.3%

% O = 49.4%

3. How many grams of iron are in 125g of iron (III) carbonate? ____47.9 g________

4. What mass of iron(III) carbonate contains 10.0g of iron? ____26.1 g________

PERCENT COMPOSITION Name:

Determine the percent composition of each of the compounds below:

1. KMnO4

% K = __24.7%_____

% Mn = __34.7%_____

% O = ___40.5%_____

2. HCl

% H = ___2.7%______

% Cl = ___97.3%_____

3. Mg(NO3)2

% Mg = ___16.4%____

% N = __18.9%______

% O = __64.7%______

4. (NH4)3PO4

% N = __28.2%____

% H = __8.1%_______

% P = ___20.8%______

% O = __43.0%______

5. Al2(SO4)3

% Al = __15.8%______

% S = ___28.1%______

% O = ___56.1%_____

Solve the following problems:

1. How many grams of oxygen can be produced from the decomposition of 100g of KClO3?

100 g KClO3 39.2 g O = 39.2 g O

100 g KClO3

2. How much iron can be recovered from 25.0g of Fe2O3?

25 g Fe2O3 69.9 g Fe = 17.5 g Fe

100 g Fe2O3

3. How much silver can be produced from 125g of Ag2S?

125 g Ag2S 87.1 g Ag = 108.9 g Ag

100 g Ag2S

HOMEWORK: PRACTICE CONVERSIONS Name:

Date:

1. Write the formula for each compound below. Then determine its molar mass.

| |FORMULA |MOLAR MASS |

|calcium sulfate |CaSO4 |136.2 g/mol |

|aluminum cyanide |Al(CN)3 |105 g/mol |

|phosphorus triiodide |PI3 |411.7 g/mol |

2. Use the molar masses you found in # 1 to make the following conversions:

a. Find the mass of 1.25moles of calcium sulfate.

1.25 moles 136.2 g = 170.3 g

1 mole

b. How many moles is 2.50g of aluminum cyanide?

2.5 g 1 mole = 0.024 moles

105 g

c. Find the mass of 0.750moles of phosphorus triiodide.

0.750 moles 411.7 g = 308.8 g

1 mole

d. How many molecules is 12.50g of phosphorus triiodide

12.50g 1 mole 6.02x1023 molecules = 1.8 x 1022 molecules

412 g 1 mole

e. Find the number of formula units in 10.0g of calcium sulfate.

10.0 g 1 mole 6.02x1023 F.U. = 4.4 x 1022 F.Un.

136 g 1 mole

MIXED MOLE CONVERSIONS:

1. Convert the quantities below to moles:

a. 14.0g of lead 1 mole __0.07 moles___________

207.2 g Pb

b. 20.5g of lithium hydroxide 1 mole ___0.86 moles___________

23.9 g LiOH

c. 14.0L of oxygen gas at STP 1 mole ___0.63 moles___________

22.4 L

2. Find the mass of:

a. 10.0L of hydrogen gas at STP 1 mole 2g H2 ____0.89 g H2____________

22.4 L 1 mole

b. 3.21 x 1022 molecules of carbon tetrachloride 154 g CCl4 ___8.2 g CCl4____________

6.02 x 1023 mlc

3. Find the empirical formula of each compound below:

a. C4H8 b. C12H24O12 c. Si3N4 d. C4Cl10

CH2 CH2O Si3N4 C2Cl5

4. Find the molecular formula of a compound if its empirical formula is C3H8, and its molar mass is

132 g/mol.

C3H8 = 44 g/Mol

132/44 = 3 ( C9H24

5. Find the molecular formula of a compound if its empirical formula is N2O5, and its molar mass is 216 g/mol.

N2O5 = 108 g/mol

216 / 108 = 2 ( N4O10

PER CENT SUGAR IN BUBBLE GUM NAME:

DATE:

PERIOD:

BACKGROUND: In this lab you are going to design and carry out an experiment to determine the percentage of sugar in bubble gum. To do so, you will assume that all of the flavoring removed from bubble gum during chewing is sugar. The chemical formula for sucrose, the sugar in bubble gum, is C12H22O11. After determining the percentage of sugar in bubble gum, you will then determine the mass of carbon and the number of carbon atoms in the sugar in a piece of bubble gum.

PROCEDURE: Write out the procedure you will follow in the space below. Include all details.

DATA: Set up a data table in the space below to record all necessary data.

CALCULATIONS: Show and explain all work. Calculations must be neat and easy to follow.

a. mass of sugar in your piece of gum: (show calculation!)

c. Mass of gum alone (without wrapper) (show calculation)

d. average % sugar in bubble gum (show calculation based on your data!)

e. % carbon in sugar (C12H22O11) (show calculation – use your class notes on % composition)

f. mass of carbon in the sugar in your piece of bubble gum (show calculation, use class notes)

(HINT: How many grams of sugar were in your gum? What per cent of sugar is carbon?)

g. atoms of carbon from the grams you calculated in “d”. (show calculation, g carbon to atoms)

CONCLUSIONS:

1. What did you learn from this experiment?

2. List two sources of error and describe the effect each had on your results. (Would the error make the per cent sugar reported too high or too low and why).

a.

b.

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Chapter 10: MOLAR MASS & PERCENT COMPOSITION

How many is a mole? 6.02 x 1023 atoms or molecules or formula units = 1 mole

How heavy is a mole? Molar mass in grams = 1 mole

How much space does a mole occupy? Gas at STP 22.4L = 1 mole

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