CALCULATING MOLAR MASS OF COMPOUNDS
a. sodium hydroxideFORMULA: ______NaOH______
Na = 23.0g
O = 16.0g
H = 1.0g
23.0 + 16.0 + 1.0 = 40.0 g/mol NaOH
b. calcium cyanideFORMULA: ____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 phosphateFORMULA: _____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) chromateFORMULA: ___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
- 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
- Find the per cent composition of potassiumsulfate:
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:
- 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
- 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
- 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 gSn(SO4)2 250 g
x = 95.5 g Sn(SO4)2
MOLECONVERSIONS
1. Find the molar mass of each compound below:
- diphosphorus pentoxide__142 g/mol ______
P2O5
- copper (II) sulfate__159.6 g/mol ______
CuSO4
- calcium cyanide___92.1 g/mol ______
Ca(CN)2
- 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)
- Find the mass of 4.50moles of diphosphorus pentoxide.______
4.5 moles 142 g = 639 g
1 mole
- How many moles is 250.0g of copper (II) sulfate?______
250.0 g 1 mole =1.57 moles
159.5 g
- 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 g1 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
- 575g of sodium sulfate to moles______
Na2SO4 (2)(23) + 32 + (4)(16) = 142 g/mole
575 g 1 mole = 4.05 moles
142 g
- 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
- 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
- 8.02 x 1023 molecules of carbon disulfide to grams______
CS2 12 + (2)(32.1) = 76.2 g/mole
8.02 x 1023molecules 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.
- 14.0L of nitrogen gas at STP to moles
14.0 L N21 mole=0.625 moles
22.4 L
- 2.5g of chlorine gas at STP to molecules
2.5 g Cl2 1 mole6.02 x 1023 molecules =2.1 x 1022 molecules
71.0 g Cl2 1 mole
- 2.24 x 1025 atoms of neon at STP to liters
2.24 x 1025 atoms Ne1 mole22.4 L=833 L
6.02 x 1023 atoms1 mole
- 13.3L of fluorine gas at STP to grams
13.3 L F21 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%_____
- 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. P4O10b. C6H12O6c. C3H6O
P2O5CH2OC3H6O
- 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
- 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 HOMEWORKNAME:
DATE:
1. Label each compound below as ionic or molecular. Write the formula and determine the molar mass of each compound.
COMPOUND / IONIC/MOLECULAR? / FORMULA / MOLAR MASS
barium sulfate / I / BaSO4 / 233.4 g/mol
tricarbon octahydride / C / C3H8 / 44 g/mol
iron (III) carbonate / I / Fe2(CO3)3 / 291.6 g/mol
tetraphosphorus decoxide / C / P4O10 / 284 g/mol
strontium phosphate / I / Sr3(PO4)2 / 452.8 g/mol
2. Find the per cent composition of iron (III) carbonate.
% Fe = 38.3%
% C = 12.3%
% O = 49.4%
- How many grams of iron are in 125g of iron (III) carbonate?____47.9 g______
- What mass of iron(III) carbonate contains 10.0g of iron?____26.1 g______
PERCENT COMPOSITIONName:
Determine the percent composition of each of the compounds below:
- KMnO4
% K = __24.7%_____
% Mn = __34.7%_____
% O = ___40.5%_____
- HCl
% H = ___2.7%______
% Cl = ___97.3%_____
- Mg(NO3)2
% Mg = ___16.4%____
% N = __18.9%______
% O = __64.7%______
- (NH4)3PO4
% N = __28.2%____
% H = __8.1%______
% P = ___20.8%______
% O = __43.0%______
- Al2(SO4)3
% Al = __15.8%______
% S = ___28.1%______
% O = ___56.1%_____
Solve the following problems:
- 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
- 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
- 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:
- 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 moles136.2 g=170.3 g
1 mole
b. How many moles is 2.50g of aluminum cyanide?
2.5 g1 mole=0.024 moles
105 g
c. Find the mass of 0.750moles of phosphorus triiodide.
0.750 moles411.7 g=308.8 g
1 mole
d. How many molecules is 12.50g of phosphorus triiodide
12.50g1 mole6.02x1023 molecules=1.8 x 1022 molecules
412 g1 mole
e. Find the number of formula units in 10.0g of calcium sulfate.
10.0 g1 mole6.02x1023 F.U. =4.4 x 1022 F.Un.
136 g1 mole
MIXED MOLE CONVERSIONS:
- Convert the quantities below to moles:
- 14.0g of lead1 mole__0.07 moles______
207.2 g Pb
b. 20.5g of lithium hydroxide1 mole___0.86 moles______
23.9 g LiOH
c. 14.0L of oxygen gas at STP1 mole___0.63 moles______
22.4 L
2. Find the mass of:
a. 10.0L of hydrogen gas at STP1 mole2g H2____0.89 g H2______
22.4 L1 mole
b. 3.21 x 1022 molecules of carbon tetrachloride154 g CCl4___8.2 g CCl4______
6.02 x 1023 mlc
3. Find the empirical formula of each compound below:
- C4H8b. C12H24O12c. Si3N4d. C4Cl10
CH2CH2OSi3N4C2Cl5
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 GUMNAME:
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!)
- Mass of gum alone (without wrapper) (show calculation)
- average % sugar in bubble gum (show calculation based on your data!)
- % carbon in sugar (C12H22O11) (show calculation – use your class notes on % composition)
- 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?)
- atoms of carbon from the grams you calculated in “d”. (show calculation, g carbon to atoms)
CONCLUSIONS:
- What did you learn from this experiment?
- 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.
1