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NOTES – MOLAR MASS AND MOLE CONVERSIONS - convert moles to liters calculator

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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 = 43.55%
% C = 24.0g/92.1g x 100 = 26.05%
% N = 28.0g/92.1g x 100 = 30.40%
% Ca = __43.55%________
% C = __26.05%________
% N = __30.40%________

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 = 17.41 g Ca

b. Find the per cent composition of magnesium phosphate:

Formula: _____ Mg3(PO4)2_____________________

% Mg = (3 x 24.3g)/262.9g x 100 = 27.73%
% P = (2 x 31.0g)/262.9g x 100 = 23.58%
% O = (8 x 16.0g)/262.9g x 100 = 48.69%
% Mg = _27.73%_________
% P = __23.58%________
% O = _48.69%_________

How many grams of magnesium are in 350g of magnesium phosphate? ___________________

27.73 g Mg = x
100 g Mg3(PO4)2 350.0 g

x = 97.06 g Mg

What mass of magnesium phosphate contains 15g of magnesium? ___________________

27.73 g Mg = 15.0 g Mg
100 g Mg3(PO4)2 x

x = 54.09 g Mg3(PO4)2

TRY:

1. Find the molar mass of each compound below:

a. lithium carbonate ____________________
FORMULA: LiCO3
Li = 6.9g
C = 12.0g
O = 3 (16.0g) = 48.0g

6.9 + 12.0 + 48.0 = 66.91 g/mol LiCO3
b. calcium nitrate _____________________

FORMULA: Ca(NO3)2
Ca = 40.0g
N = 2 (14.0g) = 28.0 g
O = 6 (16.0g) = 96.0 g

40.0 + 28.0 + 96.0 = 164.0 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 = 38.18%
% S = (2 x 31.0g)/310.9g x 100 = 19.94%
% O = (8 x 16.0g)/310.9g x 100 = 41.17%

% Sn = _38.18%_________
% S = _19.94%_________
% O = _41.17%_________

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 = 95.45 g Sn(SO4)2
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 | | | |
| |M |C3H8 |44 g/mol |
|tricarbon octahydride | | | |
| |I |Fe2(CO3)3 |291.6 g/mol |
|iron (III) carbonate | | | |
| |M |P4O10 |284.0 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 = 12.4%

% O = 49.4%

1. How many grams of iron are in 125g of iron (III) carbonate? ____47.8 g________

2. 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 = __34.8%_____
% O = ___40.5%_____

2. HCl
% H = ___2.7%______
% Cl = ___97.3%_____

3. Mg(NO3)2
% Mg = ___16.2%____
% N = __18.9%______
% O = __64.9%______

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:

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

3x16 = 39.2% 39.2 % =39.2 g O in 100g KClO3
122.6

7. How much iron can be recovered from 25.0g of Fe2O3?
MM=196.6g/mol
69.9 = x
100 25 x = 17.5 g Fe

8. How much silver can be produced from 125g of Ag2S? MM=425.3g/mol

92.5g Ag = x
100g Ag2S 125 x = 115.6 g Ag
MOLE CONVERSIONS
1. Find the molar mass of each compound below:
a. diphosphorus pentoxide __142 g/mol ____________
P2O5

b. copper (II) sulfate __159.5 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 (molecular compounds)
6.02 x 1023 formula units = 1 mole (ionic conpounds)
molar mass in grams = 1 mole (use 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 = 50.14 g
1 mole

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

110 g 1 mole 6.02 x 1023 molecules = 3.1 x 1023 molecules
213 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) = 206.5 g/mole

0.025 moles 206.5 g = 5.16 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) = 76 g/mole

8.02 x 1023 molecules 1 mole 76 g = 101.20g
6.02 x 1023 molecules 1 mole

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 g/mol |
|aluminum cyanide |Al(CN)3 |105 g/mol |
|phosphorus triiodide |PI3 |412 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 g = 170 g
1 mole

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

2.5 g 1 mole = 0.02 moles
105 g

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

0.750 moles 412 g = 309 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.U.
136 g 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. 5 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

3. 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

4. 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

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
-----------------------
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

How to go from moles to liters? M to mol/L M to mol/L Conversion. The abbreviation for M and mol/L is molar [m] and mole per liter respectively. ... Molar [M] to mol/L. Check our Molar [M] to mol/L converter and click on formula to get the conversion factor. ... M to Mole per Liter. The formula used to convert M to Mole per Liter is 1 Molar [M] = 1 Mole per Liter. ... Convert M to mol/L. ...