c20 review unit 02 chemical reactions

Chemistry 20 – Review Unit 2Chemical Reactions

- A chemical reaction can be differentiated from a phase change or a nuclear change in the following manner:

Type of Change What Changes? Example

Phase

Chemical

Nuclear

state of matter C25H52(s) → C25H52(l)

chemical composition

C25H52(s) + O2(g) → CO2(g) + H2O(g)

type of element nHeHH1

0

4

2

3

1

2

1+→+

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- In all chemical reactions, you start with one or more substances called reactants and produce substances called products.

- The plus symbol (+) is used to separate lists of chemicals and is interpreted as the word “ and ”.

- Reactants and products are separated by an arrow (→) and is interpreted as the word “ yields ”.

- How do chemists recognize that a chemical reaction has taken place? What is the evidence for a chemical reaction?

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1) Energy Exchange – 1) Energy Exchange – heat is given off (exothermic reaction) or heat is absorbed by the reactants (endothermic reaction)

Examples:

- the carbon in wood burns, releasing energy in the form of light and heat, as products are made

- C(s) + O2(g) → CO2(g) + heat + light

- exothermic reaction

- electricity ( energy ) is added to water in a process called electrolysis

- 2 H2O(l) + energy → 2 H2(g) + O2(g)

- endothermic reaction

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2) Gas (bubbles) Produced – sometimes bubbling can be easily noticed (fizzing); at other times a vapour is produced (smoke)

Examples:

- an acid is poured into a metal

- Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g)

- baking soda fizzes in water

- NaHCO3(s) + H2O(l) → CO2(g) + NaOH(aq)

3) Formation of a Precipitate – a precipitate is a solid, usually formed by mixing two clear solutions

Examples:

- two clear solutions are mixed and a yellow solid forms

- 2 NaI(aq) + Pb(NO3)2(aq) → 2 NaNO3(aq) + PbI2(s)

- iron rusts when left out in the rain

- 2 Fe(s) + 2 H2O(l) + O2(g) → 2 Fe(OH)2(s)

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4) Colour Change – a substance formed may exhibit a new characteristic colour

Examples:

- when nitric acid reacts with copper metal, a dark brown gas is produced and the resulting solution turns blue

- Cu(s) + HNO3(aq) → NO2(g) + Cu(NO3)2(aq) + H2O(l)

Note: This equation is not balanced. Technically, the equation is written differently in Chemistry 30, but for illustrative purposes, we’ll use it in this form.

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Types of Chemical Reactions

Simple Composition

- Just as the name implies, this reaction is the simplest of all. It implies that a product is composed from the simplest substances.

- A generic equation can be used to describe a simple composition reaction.

- Remember that reactants are placed on the left side of the equation and the products are found on the right side.

- element + element → compound

any number of elements

only one compound formed

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Examples

1) Word Equation

Chemical Equation (Formula)

hydrogen and oxygen yields water vapour

H2(g) + O2(g) → H2O(g)

- Is this equation correct? What is missing?

- Whenever we write out a chemical formula, we must make sure that the equation is balanced.

- The number of atoms of each element must be the same number on both reactant and product sides.

- To finish balancing the equation:

H2(g) + O2(g) → H2O(g)

- There are 2 hydrogens on the left and 2 hydrogens on the right, therefore the hydrogens are balanced.

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H2(g) + O2(g) → H2O(g)

- There are 2 oxygens on the left but 1 oxygen on the right.

- To balance the oxygens we place a number in front of water to make the oxygens balance. The number being placed is called a coefficient.

- We never place a subscript on atoms to balance, since subscripts would change the chemical composition of a reactant or product.

2 H2(g) + O2(g) → 2 H2O(g)

and not

H2(g) + O2(g) → H2O2(g)

Note that the chemical composition of water has been changed so that we do not have water any more but another chemical called hydrogen peroxide.

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2) Word Equation

Chemical Formula

calcium and chlorine yield calcium chloride

Ca(s) + Cl2(g) → CaCl2(s)

3) Word Equation

Chemical Formula

magnesium and oxygen yield magnesium oxide

Mg(s) + O2(g) → MgO(s)

Mg(s) + O2(g) → 2 MgO(s) 2 Mg(s) + O2(g) → 2 MgO(s)

4) Word Equation

Chemical Formula

aluminum and sulfur yield aluminum sulfide

Al(s) + S8(s) → Al2 S3(s)

Al(s) + S8(s) → 8 Al2 S3(s) 16 Al(s) + S8(s) → 8 Al2 S3(s) 16 Al(s) + 3 S8(s) → 8 Al2 S3(s) *

Simple Decomposition

- Simple decomposition is exactly the same as simple composition except that it is the reverse equation.

- The generic equation can be stated as follows:

compound → element + element + …

Examples

1) Potassium permanganate is decomposed.

KMnO4(s) → K(s) + Mn(s) + O2(g)

KMnO4(s) → K(s) + Mn(s) + 2 O2(g)

2) Ammonia is decomposed.

NH3(g) → N2(g) + H2(g) NH3(g) → N2(g) + 3 H2(g) 2 NH3(g) → N2(g) + 3 H2(g)

3) Calcium carbonate is decomposed.

CaCO3(s) → Ca(s) + C(s) + O2(g) CaCO3(s) → Ca(s) + C(s) + 3 O2(g) 2 CaCO3(s) → Ca(s) + C(s) + 3 O2(g) *

4) Sucrose is decomposed.

C12H22O11(s) → C(s) + H2(g) + O2(g)

C12H22O11(s) → 12 C(s) + H2(g) + O2(g)

C12H22O11(s) → 12 C(s) + 11 H2(g) + O2(g)

C12H22O11(s) → 12 C(s) + 11 H2(g) + 5.5 O2(g)

2 C12H22O11(s) → 24 C(s) + 22 H2(g) + 11 O2(g)

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Single Replacement

- As the name implies, something that is single is replaced. The generic equation is:

element + compound → elementnew + compoundnew

Example

1) sodium metal and a lithium chloride solution yield lithium metal and a sodium chloride solution

Na(s)Na(s) + LiCl(aq) →Na(s) + LiCl(aq) → Li(s)Na(s) + LiCl(aq) → Li(s) + NaCl(aq)

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How does this work?

- Break up each atom to show what their ions look like.

- Replace positive or negative atoms (switch) and reform new products.

Na(s) + LiCl(aq) → ? + ?

Na+Na+ Li + Cl −Na+ Li + Cl − Li + Na+ Cl −

Na(s) +Na(s) + LiCl(aq) →Na(s) + LiCl(aq) → Li(s) +Na(s) + LiCl(aq) → Li(s) + NaCl(aq)

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Double Replacement

- In double replacement reactions, two compounds are reacted to form two new compounds.

- Generic formula:

compound1 + compound2 → compound3 + compound4

Examples

1) Sodium chloride solution and silver nitrate solutions are mixed.

NaCl(aq) +NaCl(aq) + AgNO3(aq) →NaCl(aq) + AgNO3(aq) → ? + ?

Na+ Cl −Na+ Cl − Ag+Na+ Cl − Ag+ NO3− →Na+ Cl − Ag+ NO3− → forms new products

join

Na+ Cl − Ag+ NO3− → Na+ NO3

− +

join

Na+ Cl − Ag+ NO3− → Na+ NO3

− + Ag+ Cl -

NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)

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2) Solutions of lead (II) nitrate and sodium iodide are mixed.

Pb(NO3)2(aq) +Pb(NO3)2(aq) + NaI(aq) →Pb(NO3)2(aq) + NaI(aq) → ? + ?

Pb2+ NO3– +Pb2+ NO3– + Na+ I – →Pb2+ NO3– + Na+ I – → Pb2+ I – +Pb2+ NO3– + Na+ I – → Pb2+ I – + Na+ NO3

–

Pb(NO3)2(aq) +Pb(NO3)2(aq) + NaI(aq) →Pb(NO3)2(aq) + NaI(aq) → PbI2(s) +Pb(NO3)2(aq) + NaI(aq) → PbI2(s) + NaNO3(aq)

Pb(NO3)2(aq) + NaI(aq) → PbI2(s) + 2 NaNO3(aq) Pb(NO3)2(aq) + 2 NaI(aq) → PbI2(s) + 2 NaNO3(aq)

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3) Solutions of gallium chloride and hydrosulfuric acid are mixed.

GaCl3(aq) +GaCl3(aq) + H2 S(aq) → ? + ?

Ga 3+ Cl – +Ga 3+ Cl – + H + S 2– →Ga 3+ Cl – + H + S 2– → Ga 3+ S 2– +Ga 3+ Cl – + H + S 2– → Ga 3+ S 2– + H + Cl –

GaCl3(aq) +GaCl3(aq) + H2 S(aq) →GaCl3(aq) + H2 S(aq) → Ga2 S3(aq) +GaCl3(aq) + H2 S(aq) → Ga2 S3(aq) + HCl(aq)

_ GaCl3(aq) + _ H2 S(aq) → _ Ga2 S3(aq) + _ HCl(aq) _ GaCl3(aq) + 3 H2 S(aq) → _ Ga2 S3(aq) + _ HCl(aq) _ GaCl3(aq) + _ H2 S(aq) → _ Ga2 S3(aq) + 6 HCl(aq) 2 GaCl3(aq) + _ H2 S(aq) → _ Ga2 S3(aq) + _ HCl(aq) _ GaCl3(aq) + _ H2 S(aq) → 1 Ga2 S3(aq) + _ HCl(aq)

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4) Solutions of calcium nitrate and sodium phosphate are mixed.

Ca(NO3)2(aq) + Na3PO4(aq) → ? + ?

Ca2+ NO3– +Ca2+ NO3– + Na+ PO4

3– →Ca2+ NO3– + Na+ PO4

3– → Ca2+ PO43– +Ca2+ NO3

– + Na+ PO43– → Ca2+ PO4

3– + Na+ NO3–

Ca(NO3)2(aq) +Ca(NO3)2(aq) + Na3PO4(aq) →Ca(NO3)2(aq) + Na3PO4(aq) → Ca3(PO4)2(aq) +Ca(NO3)2(aq) + Na3PO4(aq) → Ca3(PO4)2(aq) + NaNO3(aq)

_ Ca(NO3)2(aq) + _ Na3PO4(aq) → _ Ca3(PO4)2(s) + _ NaNO3(aq) 3 Ca(NO3)2(aq) + _ Na3PO4(aq) → _ Ca3(PO4)2(s) + _ NaNO3(aq) _ Ca(NO3)2(aq) + _ Na3PO4(aq) → _ Ca3(PO4)2(s) + 6 NaNO3(aq) _ Ca(NO3)2(aq) + 2 Na3PO4(aq) → _ Ca3(PO4)2(s) + _ NaNO3(aq) _ Ca(NO3)2(aq) + _ Na3PO4(aq) → 1 Ca3(PO4)2(s) + _ NaNO3(aq)

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Hydrocarbon Combustion

In hydrocarbon combustion, a fuel made up of at least hydrogen and carbon, is burned (combustion) in the presence of oxygen.

Generic Formula:

hydrocarbon +hydrocarbon + O2(g) →hydrocarbon + O2(g) → CO2(g) +hydrocarbon + O2(g) → CO2(g) + H2O(g)

Examples

1) Methane is burned to heat our homes in winter.

CH4(g) +CH4(g) + O2(g) →CH4(g) + O2(g) → CO2(g) +CH4(g) + O2(g) → CO2(g) + H2O(g)

_ CH4(g) + _ O2(g) → _ CO2(g) + _ H2O(g) _ CH4(g) + _ O2(g) → 1 CO2(g) + _ H2O(g) _ CH4(g) + _ O2(g) → _ CO2(g) + 2 H2O(g) _ CH4(g) + 2 O2(g) → _ CO2(g) + _ H2O(g)

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2) Gasoline is combusted in a car engine.

C8H18(l) +C8H18(l) + O2(g) → CO2(g) + H2O(g)

_ C8H18(l) + _ O2(g) → _ CO2(g) + _ H2O(g) _ C8H18(l) + _ O2(g) → 8 CO2(g) + _ H2O(g) _ C8H18(l) + _ O2(g) → _ CO2(g) + 9 H2O(g) 2

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2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)

3) The burning of wood alcohol produces a flame.

CH3OH(l) +CH3OH(l) + O2(g) → CO2(g) + H2O(g)

_ CH3OH(l) + _ O2(g) → _ CO2(g) + _ H2O(g) _ CH3OH(l) + _ O2(g) → 1 CO2(g) + _ H2O(g) _ CH3OH(l) + _ O2(g) → _ CO2(g) + 2 H2O(g) 2

3

2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(g)

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4) We burn glucose in our body cells.

C6H12O6(s) +C6H12O6(s) + O2(g) → CO2(g) + H2O(g)

_ C6H12O6(s) + _ O2(g) → _ CO2(g) + _ H2O(g) _ C6H12O6(s) + _ O2(g) → 6 CO2(g) + _ H2O(g) _ C6H12O6(s) + _ O2(g) → _ CO2(g) + 6 H2O(g) _ C6H12O6(s) + 6 O2(g) → _ CO2(g) + _ H2O(g)

Other

Some chemical reactions do not fit in any of the previously studied chemical reactions. These reactions will then be placed in a category called “other”.

Examples

1) sulfur trioxide and water → sulfuric acid

SO3(g) +SO3(g) + H2O(l) →SO3(g) + H2O(l) → H2SO4(aq)

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2) Nitrogen monoxide gas reacts with oxygen to produce nitrogen dioxide gas.

NO(g) + NO(g) + O2(g) →NO(g) + O2(g) → NO2(g)

_ NO(g) + _ O2(g) → _ NO2(g) 2 NO(g) + _ O2(g) → _ NO2(g) _ NO(g) + _ O2(g) → 2 NO2(g)

3) A sodium bicarbonate solution decomposes into a sodium carbonate solution, carbon dioxide and water.

NaHCO3(aq) →NaHCO3(aq) → Na2CO3(aq) +NaHCO3(aq) → Na2CO3(aq) + CO2(g) +NaHCO3(aq) → Na2CO3(aq) + CO2(g) + H2O(l)

_ NaHCO3(aq) → _ Na2CO3(aq) + _ CO2(g) + _ H2O(l) 2 NaHCO3(aq) → _ Na2CO3(aq) + _ CO2(g) + _ H2O(l)

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Determining Whether a Substance Dissolves Easily In Water or Forms a Precipitate (Solid)

- We are told that ionic substances are solids at room temperature and molecular substances can be any state depending on which ones we are talking about.

- What happens if we try to dissolve them in water? Will they easily dissolve or will they tend not to dissolve, thereby remaining a solid?

- In chemical reactions where solutions are involved, it is necessary to know if a substance dissolves, so that we can place (aq) after its chemical formula.

- To determine the “solubility” of a molecular substance we will need to memorize which ones easily dissolve and which ones don’t.

- To determine the “solubility” of an ionic substance we will use the “Solubility Table” found at the bottom of the Periodic Table of Ions Sheet.

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How does this work?

- Suppose you want to know if an ionic substance such as salt, NaCl(s),

dissolves in water.

- First, find the ions that make up the compound. [ Na + and Cl – ]

- Next, locate one of those ions in the top row. [ Na + is a Gr IA ion ]

- What does “All” and “None” mean?

- Any Gr IA ion combined with “All” other negative ions will be “Soluble”, therefore we we put (aq) after that compound’s chemical formula. [ NaCl(aq) ]

Which of the following ionic compounds will be soluble?

a) ammonium hydroxide

b) calcium phosphate

c) potassium hydroxide

d) strontium carbonate

e) silver chloride

NH4OH(aq)

Ca3(PO4)2(s)

KOH(aq)

SrCO3(s)

AgCl(s)

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Predict the states of matter for each of the following chemical reactions.

1) Crystals of sodium iodide and lead (II) nitrate are placed in separate beakers containing distilled water. Then their contents are mixed. Write out the balanced chemical equation.

_ NaI(aq) +_ NaI(aq) + _ Pb(NO3)2(aq) →_ NaI(aq) + _ Pb(NO3)2(aq) → _ NaNO3(aq) +_ NaI(aq) + _ Pb(NO3)2(aq) → _ NaNO3(aq) + _ PbI2(s)2 NaI(aq) + _ Pb(NO3)2(aq) → _ NaNO3(aq) + _ PbI2(s)_ NaI(aq) + _ Pb(NO3)2(aq) → 2 NaNO3(aq) + _ PbI2(s)

2) Crystals of silver nitrate and sodium chloride are placed in separate beakers containing distilled water. Then their contents are mixed. Write out the balanced chemical equation.

_ AgNO3(aq) +_ AgNO3(aq) + _ NaCl(aq) →_ AgNO3(aq) + _ NaCl(aq) → _ AgCl(s) +_ AgNO3(aq) + _ NaCl(aq) → _ AgCl(s) + _ NaNO3(aq)

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Solubility Table For Ionic Substances

Ion H + Gr IA NH4+ NO3

– CH3COO –

Cl – Br –

I –

SO42– S2– OH –

PO43 –

SO32 –

CO32 –

Soluble(aq)

All All All All All Most MostGr IAGr IIANH4

+

Gr IANH4

+

Sr2+

Ba2+

Gr IANH4

+

LowSolubility

(s)None None None None None

Ag+

Pb2+

Hg+

Cu+

Ag+

Pb2+

Ca2+

Ba2+

Sr2+

Most Most Most

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The Mole

Before we discuss the concept of the “mole”, it is important to be able to work with numbers.

Scientific Notation

Example: Convert 4638600000 into scientific notation

4.638600000 x 10 9

Example: Convert 0.00000475 into scientific notation

4.75 x 10 –6

Note: A positive exponent indicates a number greater than 1 and a negative exponent indicates a number smaller than 1.

Example: Calculate the following.

a) (4.50 x 103) (9.25 x 105) =

b) (3.36 x 10–7) (5.50 x 104) =

c) =−

−

5

4

10x8.24

10x7.73

4.16 x 10 9

1.85 x 10 –2

9.38 26

d) =−

−

)10x)(7.2710x(2.58

)10x)(2.3410x(5.5554

59

e) = )10x)(4.7110x(8.22

)10x)(2.3410x(7.12x

)10x)(2.3310x(1.48

)10x)(9.1110x(3.5254

56

54

53

−−

−

f) =)10x)(4.7110x(8.22

)10x)(2.3410x(7.12

)10x)(2.3310x(1.48

)10x)(9.1110x(3.5254

56

54

53

−−

−

+

6.92 x 10 –4

4.00 x 10 9

4.30 x 10 19

Significant Digits

- The number 354 has _ significant digits. - The number 354 has 3 significant digits.

- The number 35.4 has _ significant digits. - The number 35.4 has 3 significant digits.


- The number 3.54 x 104 has _ significant digits. - The number 3.54 x 104 has 3 significant digits.





- The number 0.00350040 has _ significant digits. - The number 0.00350040 has 6 significant digits. 27

Rules For Multiplying and Dividing

- When multiplying or dividing, the answer must contain the least number of significant digits that are found in the numbers being calculated.

- Example: 3.3 x 0.134 = - The answer is 0.4422 but since the least number of significant digits

being used is “_”, the answer must be rounded off to “_” digits. - The answer is 0.4422 but since the least number of significant digits

being used is “2”, the answer must be rounded off to “2” digits.

Answer:Answer: 0.44

- Example: 3746 x 0.120 =

- The calculator answer is 449.52, but since the least number of significant digits is “_”, the answer is rounded off to _ digits.

- The calculator answer is 449.52, but since the least number of significant digits is “3”, the answer is rounded off to 3 digits.

- Example: 3746 x 120 =

- The calculator answer is 449520, but since the least number of significant digits is “_”, the answer is rounded off to _ digits.

- The calculator answer is 449520, but since the least number of significant digits is “3”, the answer is rounded off to 3 digits.

- The answer must be about 450000, but that number has _ digits. - The answer must be about 450000, but that number has 6 digits.

- In this case, use scientific notation. Answer:Answer: 4.50 x 10 5

Answer:Answer: 450

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Rules for Adding and Subtracting

- Perform the calculation in your calculator.

- Determine the least number of place holding there are after the decimal in each entry.

- Your answer must contain the least number of place holdings after the decimal.

- Round off the answer.

Example: 34.65 + 4.235 =

- Calculator answer is- Calculator answer is 38.885. - The least number of place holding after the decimal is “_”. - The least number of place holding after the decimal is “2”.

Answer: Answer: 38.89

- Example: 1356.245 + 245.33 – 0.0001 = 1601.5749 = 1601.5749 = 1601.57

Note: When calculating long series of numbers, round off only at the end.

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The Mole

- Atoms are extremely small so it becomes very difficult to make measurements when only a few atoms or molecules are used.

- If we wanted to take the mass of 1 atom, it would be appropriate to find the mass of say, 1000 atoms and then divide by 1000, but 1000 atoms still gives us a very small mass to measure.

- What number of atoms or molecules would be large enough? 1 dozen? 1 gross (144)? 1 million?

- In the early 1800’s, a scientist named Amedeo Avogadro devised a way of measuring the masses of very small particles of matter.

- Avogadro suggested that if you took exactly 12.00 g of the carbon-12 atom you would have a very large number of carbon atoms. Just as words like “dozen”, “gross” or “couple” have numerical meaning, he needed a word to describe how many atoms there would be in 12.00 g of carbon-12.

- This word is now known as the “mole”.

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- One dozen atoms would be 12 atoms.

- One gross of atoms would be 144 atoms.

- One mole of atoms is 6.02 x 1023 atoms, now known as Avogadro’s number.

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The Green Pea Analogy (from Alchem 2000)

- One hundred (102) green peas have a volume of about 20 cm3 or 20 mL in a graduated cylinder.

- One million (106) green peas have enough volume to fill an ordinary refrigerator.

- One billion (109) green peas can fill an average 3 bedroom house.

- One trillion (1012) green peas can fill 1000 average homes.

- One quadrillion (1015) green peas can fill all the building in a larger city such as Edmonton.

- Imagine all of Alberta covered one metre deep in green peas. Now you got one quintillion (1018) green peas.

- Imagine that all the continents are now covered 1 metre deep with green peas. Now you have one sextillion (1021).

- If you froze all the oceans and totally covered the whole earth with green peas and had 250 planets just like earth, all covered with green peas, you would now have a mole’s worth!

- 250 000 planets like earth, all covered one metre deep in green peas would give you a cotillion (1027), the number of atoms in your body! 32

Molar Mass

- The molar mass of any element can be easily found on a Periodic Table of Elements.

- The average molar mass of all carbon isotopes is 12.01 . This

means that one mole (6.02 x 1023) of carbon isotopes has a mass of 12.01 g.

mol

g

- One mole of carbon-12 atoms has a mass of 12.00 g.

- One mole (1 mol) of sodium atoms has a mass of 22.99 g.

- One mole of “eggs” would have a mass of a heck of a lot!

- One mole of anything is 6.02 x 1023 number of things.

- 1 mol of Mg(s) atoms = 6.02 x 1023 atoms = 24.31 g

- 1 mol of N2(g) molecules = 6.02 x 1023 molecules = 28.02 g

- 1 mol of H2O(l) molecules = 6.02 x 1023 molecules = 18.02 g

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Calculating Number of Particles In Given Number of Moles

Examples:

1. How many atoms are there in 2.50 mol of lithium?

Since 1 mol of Li(s) = Since 1 mol of Li(s) = 6.02 x 10 23 atomsSince 1 mol of Li(s) = 6.02 x 10 23 atoms

then 2.50 mol of Li(s) should have

2.50 mol x 6.02 x 10 23 atoms =

Since 1 mol of Li(s) = 6.02 x 10 23 atoms

then 2.50 mol of Li(s) should have

2.50 mol x 6.02 x 10 23 atoms = 1.51 x 10 24 atoms

or

Number'sAvogadromolesofnumber#atoms ×=

molatoms10x02.6

mol50.2#atoms23

×=

atoms10x51.1#atoms 24=

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2. How many molecules are there in 25.0 mol of water?

Numbers'Avogadromolesofnumbermolecules# ×=

molmolecules10x02.6

mol0.25molecules#23

×=

molecules10x51.1molecules# 25=

3. How many moles are there in 4.55 x 1015 atoms of zinc?

( )

=

molatoms2310x02.6

atoms#mol#

( )

=

molatoms10x02.6

atoms10x55.4mol#

23

15

mol10x56.7mol# 9−=35

4. How many moles are there in 2.75 x 1024 molecules of ammonia?

( )

=

molmolecule10x02.6

molecules#mol#

23

( )

=

mol

molecules10x02.6

molecules10x75.2mol#

23

24

mol57.4mol# =

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Calculating the Molar Mass of Any Substance

Find the molar masses for each of the following examples:

1) Cu(s)

Since the formula suggests that copper is made up only one type of atom, we simply look up the molar mass value from the periodic table.

∴ Cu(s) =∴ Cu(s) = 63.55 g in one mole = 63.55 mol

g

2) H2O(l)

There are 2 hydrogens and 1 oxygen involved in making the water molecule ∴ we need to add the molar masses of all atoms.

2 – H =2 – H = 2 x 1.01 =2 – H = 2 x 1.01 = 2.02 g

1 – O =1 – O = 1 x 16.00 =1 – O = 1 x 16.00 = 16.00 g

H2O(l) =H2O(l) = 18.02 molg

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3) CaSO4(s)

1 – Ca =1 – Ca = 1 x 40.08 =1 – Ca = 1 x 40.08 = 40.08 g

1 – S =1 – S = 1 x 32.06 =1 – S = 1 x 32.06 = 32.06 g

4 – O =4 – O = 4 x 16.00 =4 – O = 4 x 16.00 = 64.00 g

CaSO4(s) =CaSO4(s) = 136.14 molg

4) Al(OH)3(s)

1 – Al =1 – Al = 1 x 26.98 =1 – Al = 1 x 26.98 = 26.98 g

3 – O =3 – O = 3 x 16.00 =3 – O = 3 x 16.00 = 48.00 g 3 – H =3 – H = 3 x 1.01 =3 – H = 3 x 1.01 = 3.03 g

Al(OH)3(s) =Al(OH)3(s) = 78.01 molg

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5) Na2H2PO4(s)

2 – Na =2 – Na = 2 x 22.99 =2 – Na = 2 x 22.99 = 45.98 g

2 – H =2 – H = 2 x 1.01 =2 – H = 2 x 1.01 = 2.02 g

1 – P =1 – P = 1 x 30.97 =1 – P = 1 x 30.97 = 30.97 g

4 – O =4 – O = 4 x 16.00 =4 – O = 4 x 16.00 = 64.00 g

Na2H2PO4(s) =Na2H2PO4(s) = 142.97 molg

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Calculations Converting Mass to Moles

We can use a simple formula for this type of calculation:

massmolar

massmolesofnumber =

Mm

n =

Examples

1. How many moles are there in 75.0 g of water?

- molar mass of water = mol

g02.18

Mm

n = ( )( )mol

g02.18

g0.75= mol16.4=

40

2. How many moles are there in 100 g of sucrose?

- molar mass of sucrose = mol

g34.342

Mm

n = ( )( )mol

g34.342

g100= mol292.0=

Calculations Converting Moles to Mass

We can use a simple formula for this type of calculation:

massmolarmolesofnumbermass ×=nMm =

41

Examples

1. What mass do you have in 2.83 mol of table salt?

nMm =

( )

=

molg

44.58mol83.2m

g165m =

2. What is the mass of 25.0 mol of ammonia?

nMm =

( )

=

molg

04.17mol0.25m

g426m =

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c20 review unit 02 chemical reactions

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