Representative Particles

Whatever type of particle that is being examined.

•Atom – any single element that is part of the overall compound.

•Ion – an atom or group of atoms that has a net charge

•Molecule – a group of atoms that have been chemically combined. There is no net charge on a molecule.

Determine how many atoms, ions and molecules are in each example:Pb(NO3)2

1 molecule

3 ions ( 1 Pb+2, 2 NO3-1)

9 atoms ( 1 Pb, 2 N, 6 O)

5 (NH4)2SO4

5 molecules

15 ions ( 10 NH4+1, 5 SO4-2)

75 atoms ( 10 N, 40 H, 5 S, 20 O)

compoundcompound AtomsAtoms IonsIons moleculesmolecules

7 NaCl7 NaCl 1414 1414 77

3 K3 K22COCO33 1818 99 33

8 AgNO8 AgNO33 4040 1616 88

5 Ca5 Ca33(PO(PO44))22 6565 2525 55

6(NH6(NH44))22COCO33 8484 1818 66

The size of an atom, ion or molecule is too small to deal with on an individual basis in the lab. Therefore, normally in chemistry we deal in moles of atoms, ions or molecules. A mole can have multiple definitions:

1 mole = 6.022 E 23 (Avogadro’s number)

= atomic weight in grams

= 22.4 liters (for gases only)

Molar Mass – the mass, in grams, of 1 mole of any substance. It is found by taking the atomic weight from the periodic table, and rounding to one place past the decimal.

The molar mass of each of the following:

Be = 9.0 grams

Cr = 52.0 grams

Cl = 35.5 grams

S = 32.1 grams

O = 16.0 grams

Calculate the molar mass of the following:BeCl2, BeSO4, Cr(ClO3)3, Cr2(SO4)3 and (NH4)3PO4

BeCl2 : 9.0 + 2 (35.5) = 80.0 grams

BeSO4 : 9.0 + 32.1 + 4 (16.0) = 105.1 grams

Cr(ClO3)3 : 52.0 + 3 (35.5) + 9 (16.0) = 302.5 grams

Cr2(SO4)3 : 2 (52.0) + 3 (32.1) + 12 (16.0) = 392.3 grams

(NH4)3PO4 : 3 (14.0) + 12 (1.0) + 31.0 + 4 (16.0) = 149.0 g

The Flux CapacitorParticles Mass

6.022 E23 molar mass

Moles

22.4

Volume

1 mole = 6.022 E23 (particles) = molar mass (grams) = 22.4 (liters)

What is the mass of 1.75 mole of NaCl?

1.75 mol NaCl 58.5 g NaCl = 102 g NaCl

1 mol NaCl

How many moles of AgNO3 is 57.3 grams?

57.3 g AgNO3 1 mol AgNO3 = 0.337 mol AgNO3

169.9 g AgNO3

What is the volume of 0.258 moles of CH4?

0.258 mol CH4 22.4 l CH4 = 5.78 liter CH4

1 mol CH4

1.How many moles of FeSO4 is 2.57 grams of FeSO4?

2.How many molecules are in 2.57 grams of FeSO4?

3.How many moles of oxygen are in 2.57 grams of FeSO4?

4.How many ions are in 2.57 grams of FeSO4?

1. 2.57 g FeSO4 1 mol FeSO4 = 0.0169 mol FeSO4

151.9 g FeSO4

2. 2.57 g FeSO4 1 mol FeSO4 6.022 E 23 molc FeSO4 =

151.9 g FeSO4 1 mol FeSO4

1.02 E22 molc FeSO4

3. 2.57 g FeSO4 1 mol FeSO4 4 mol O = 0.0677 mol O

151.9 g FeSO4 1 mol FeSO4

4. 2.57 g FeSO4 1 mol FeSO4 6.022 E 23 molc 2 ions =

151.9 g FeSO4 1 mol FeSO4 1 molc

2.04 E 22 ions

How many atoms are in 17.3 grams of Fe2O3?

17.3 g Fe2O3 1 mol Fe2O3 6.02 E 23 Fe2O3 5 atoms =

159.6 g Fe2O3 1 mol Fe2O3 1 Fe2O3

3.26 E 23 atoms

What is the mass of 1.16 E 21 molecules of Fe2O3?

1.16 E 21 molc Fe2O3 1 mol Fe2O3 159.6 g Fe2O3 =

6.02 E 23 molc Fe2O31 mol Fe2O3

0.307 g Fe2O3

1.What is the mass of 617 molecules of MgSO4?

2.What is the volume of 18.7 grams of O2?

3.How many Freon molecules are in 0.1573 liters of Freon?

4.How many oxygen atoms are in 16.3 g of CuClO3?

5.What is the volume of 76.3 grams of mercury vapor?

1.What is the volume of 16.3 g of O2?

2.How many molecules are there in 63.75 liters of CO2?

3.What is the mass of 9.47 E 25 molecules of CH4?

4.a. How many moles of PCl3 are in 16.8 grams of PCl3?

b. How many atoms are in 16.8 g of PCl3?

5. What is the mass of 27.9 liters of NH3?

1.What is the volume of 4.35 E 18 molecules of SO2?

2.How many atoms are in 15.3 liters of propane gas, C3H8?

3.What is the mass of 3.92 liters of HCCl3?

4.How many oxygen atoms are in 16.3 grams of K2Cr2O7?

5.What is the mass of 8.15 E 24 molecules of CCl4?

6.How many grams of HClO4 can be made from 9.13 E 24 oxygen atoms?

You are starting with 18.3 g of Mg(NO3)2.

1.How many moles of Mg(NO3)2 are there?

2. How many moles of N are there?

3. How many grams of oxygen are present?

1. 18.3 g Mg(NO3)2 1 mole Mg(NO3)2 = 0.123 mol Mg(NO3)2

148.3 g Mg(NO3)2

2. 18.3 g Mg(NO3)2 1 mol Mg(NO3)2 2 mol N = 0.247 mol N

148.3 g Mg(NO3)2 1 mol Mg(NO3)2

3. 18.3 g Mg(NO3)2 1 mol Mg(NO3)2 96.0 g O = 11.8 g O

148.3 Mg(NO3)2 1 mol Mg(NO3)2

% Composition - % by mass of each element in a compound

• What is the % of each element in K2CrO4?

1.Calculate the molar mass

2 (39.1) + 52.0 + 4 (16.0) = 194.2 grams

2. Take the mass of each element and divide by the molar mass of the compound, and then multiply by 100%

% K = 78.2/194.2 x 100% = 40.3 %

% Cr = 52.0/194.2 x 100 % = 26.8 %

% O= 64.0/194.2 x 100 % = 33.0 %

The total % should be between 99.8 – 100.2 %

•What is the % composition of C3H8?

3 (12.0) + 8 (1.0) = 44.0 g

% C = 36.0/44.0 x 100 % = 81.8 %

% H = 8.0/44.0 x 100 % = 18 %

•What is the % composition of NaHSO4?

23.0 + 1.0 + 32.1 + 4 (16.0) = 120.1 g

% Na = 19.2 % % S = 26.7 %

% H = 0.83 % % O = 53.3 %

How many grams of sodium are in 4.76 grams of NaHSO4?

4.76 g NaHSO4 23.0 g Na = 0.912 g Na

120.1 g NaHSO4

How many grams of oxygen are in 4.76 grams of NaHSO4?

4.76 g NaHSO4 64.0 g O = 2.54 g O

120.1 g NaHSO4

You have 16.3 grams of Ca3(PO4)2

1.What is the % composition of Ca3(PO4)2?

2.How many moles of phosphorus?

3.How many calcium atoms?

4.How many grams of oxygen?

5.How many moles of Ca3(PO4)2?

6.How many ions?

3(40.1) + 2(31.0) + 8(16.0) = 310.3 g

1.% Ca = 38.8 % % P = 20.0 % % O = 41.3 %

2.16.3 g Ca3(PO4)2 1 mol Ca3(PO4)2 2 mol P = 0.105 mol P

310.3 g Ca3(PO4)2 1 mol Ca3(PO4)2

3.16.3 g Ca3(PO4)2 1 mol Ca3(PO4)2 6.022 E23 molc 1 Ca atom = 310.3 g Ca3(PO4)2 1 mol Ca3(PO4)2 1 molc

3.16 E22 Ca atoms

4. 16.3 g Ca3(PO4)2 128 g O = 6.72 g O 310.3 g Ca3(PO4)2

5.16.3 g Ca3(PO4)2 1 mol Ca3(PO4)2 = 0.0525 mol Ca3(PO4)2

310.3 g Ca3(PO4)2

6.16.3 g Ca3(PO4)2 1 mol Ca3(PO4)2 6.022 E23 molc 5 ions = 310.3 g Ca3(PO4)2 1 mol Ca3(PO4)2 1 molc

1.58 E 23 ions

Empirical Formula – lowest ratio of elements in a compound (all ionic compounds)

Molecular Formula – Actual number of elements in a compound (usually covalent compounds)

MM = molecular massEM = empirical mass

MM/EM = whole number. Multiply subscripts of the empirical formula by the whole number to get the molecular formula.

To get the empirical formula from the molecular formula, reduce the subscripts.

What is the molecular formula of a compound with a molecular mass of 60.0 grams, and an empirical formula of CH4N?

EM = 12.0 + 4 (1.0) + 14.0 = 30.0 grams

MM/EM = 60.0/30.0 = 2

2 x (CH4N) = C2H8N2

Emp. Emp. formulaformula

MMMM

(g)(g)

EMEM

(g)(g)

MM/EMMM/EM Molc.Molc.

formulaformula

CC22HClHCl 181.5181.5

CHCH22OO 180.0180.0

CHCH 78.078.0

HOHO 34.034.0

CHCH44 16.016.0

Emp. Emp. formulaformula

MMMM

(g)(g)

EMEM

(g)(g)

MM/EMMM/EM Molc.Molc.

formulaformula

CC22HClHCl 181.5181.5 60.560.5 33 CC66HH33ClCl33

CHCH22OO 180.0180.0 30.030.0 66 CC66HH1212OO66

CHCH 78.078.0 13.013.0 66 CC66HH66

HOHO 34.034.0 17.017.0 22 HH22OO22

CHCH44 16.016.0 16.016.0 11 CHCH44