Chemistry- Chemical Equations & Compostion Of Compounds

Front 1

Mole

Back 1

Avogradro's Number
1 mole = 6.022 x 10^24

Front 2

Molar Mass

Back 2

The *atomic mass* of an element in grams containing Avogradro's number of atoms.

Front 3

Example: How many moles of iron does 25.0g of iron (Fe) represent?

SOLUTION MAP: g Fe ---> moles Fe
Given: 1 mole Fe = 56.85g (found in PERIODIC TABLE)

Back 3

(25.0 g Fe) x (1 mole Fe) / (55.85 g Fe)
= 0.448 mol Fe
(25.0g has 3 sig-figs so the number allowed in the answer has to be 3)

Front 4

Solve: How many magnesium atoms are contained in 5.00 g Mg?

SOLUTION MAP: g Mg ---> atoms Mg

Back 4

1 mole Mg = 24.31 g (PERIODIC TABLE)

Needed: 6.022 x 10^23 (Avogadro's #)

(5.00g Mg) x (6.022 x 10^23 atoms Mg) / (24.31 g Mg) = 1.24 x 10^23 atoms Mg

Front 5

Solve: What is the mass of 0.252 mol of copper (Cu)?

Back 5

SOLUTION MAP: mol Cu ---> grams Cu

1 mole Cu = 63.55gCu (PERIODIC TABLE)

(0.252 mol Cu) x (63.55g Cu) / (1 mol Cu) = 16.0g Cu
(sig figs = 3)

Front 6

Solve: How many oxygen atoms are present in 1.00 mol of oxygen molecules?

SOLUTION MAP: moles O2 ---> molecules O2 ---> atoms O
Given: 2 atoms of O = 1 molecule O2

Back 6

Needed: Avogadros # - 6.022 x 10^23

(1.00 mol O2) x (6.022 x 10^23 molecules O2) / (1 mole O2) x (2 atoms O) / (1 molecule O2) = 1.20 x 10^24 atoms O

Front 7

Finding Molar Mass

(Ex: What is the molar mass of H2O (Water))
Known: H2O
PERIODIC TABLE: 2 H = 2 x 1.008g = 2.016
1 O = 16.00g
ADD 16.00 + 2.016

Back 7

= 18.02g

Front 8

Find the Molar Mass of:

Calcium Hydroxide- Ca(OH)2

Back 8

1 Ca = 1 x 40.08g
2 O = 2 x 16.00g
2 H = 2 x 1.008g
= 74.10g molar mass of Ca(OH)2

Front 9

Find the Molar Mass of:

KNO3

Back 9

= 101.1 g KNO3

Front 10

How many moles of sodium hydroxide (NaOH) are there in 1.00kg of sodium hydroxide?

Known: 1.00 kg NaOH
SOLUTION MAP: kg NaOH --> g NaOH --> mol NaOH

*Conversion factors - 1000g/1 kg AND 1 mole NaOH / 40.00g NaOH*

Back 10

(1.00kg NaOH) x (1000g NaOH) / (1 kg NaOH) x (1 mol NaOH) / (40.00 g NaOH)
= 25.00 mol NaOH

Front 11

How many molecules of hydrogen chloride (HCI) are there in 25.0 kg of hydrogen chloride?

1 Given conversion factor- 6.022 x 10^23 molecules HCI / 1 mol HCI

Back 11

Known: 25.0g HCI
SOLUTION MAP: g HCI ---> mol HCI ---> molecules HCI

(25.0 g HCI) x (1 mol HCI) / (36.45g HCI) x (6.022 x 10^23 molecules HCI) / (1 mole HCI)
= 4.13 x 10^23 molecules HCI

Front 12

Finding Percent Composition:

(Ex: NaCI)

Percent mass - (Total Mass of the Element) / (Molar Mass) x 100

Na = 22.99g (atomic mass) (PERIODIC TABLE)
Cl = 35.45g (atomic mass) (PERIODIC TABLE)

Back 12

= 58.44 g molar mass of NaCI

Front 13

Find the percent mass of:

KCI

Back 13

K= 39.10g
CI= 35.35g
=74.55g molar mass of KCI

Front 14

Calculate the percent composition of potassium sulfate (K2SO4)

Known: K2SO4
1- Find the atomic masses of K, S, and O. Add and get the molar mass
2- Calculate percent composition of each element
3- Add up all percentages

Back 14

K: (78.20g K / 174.3g) x 100 = 44.87% K

S: (32.07g S / 174.3g) x 100 = 18.40% S

O: (64.00g O / 174.3) x 100 = 36.72% O

= 99.99% total

Front 15

Calculate the percent composition of Ca(NO3)2

Back 15

Ca: 24.42%

N: 17.07%

O: 58.50%

= 99.99% total

Front 16

Empirical Formula

Back 16

Gives the smallest whole-number ratio of atoms present in a compound.

(Ex. CH)

Front 17

Molecular Formula

Back 17

Represents the total number of atoms of each element present in one molecule of a compound.

( Ex. (CH)2 = C2H2 ) (Acetylene)

Front 18

Calculating Empirical Formulas

Back 18

1) Assume a definite starting quantity, if not given, express the mass of each element in grams
2) Convert grams of each element to moles using each element's molar mass
3) Divide each value obtained in #2 by the smallest of these values. If the numbers are whole, use them as subscripts and write the empirical formula. If not whole, go to #4.
4) Multiply the values obtained in #3 by the smallest number that will convert them to whole numbers and write them as subscripts in the empirical formula.

(Ex: if the ratio of A to B is 1:0 and 1:5, multiply both numbers by 2 to obtain a ratio of 2:3. The empirical formula = A2B3)

Front 19

Calculate the empirical formula of a compound containing 11.19% hydrogen (H) abd 88.79% oxygen (O).

Known: 11.19% H
88.79% O

1) Assume 100g of material. We know the percent of each element equals the grams of each element
2) Convert grams of each element to moles

H: 11.19g x (1 mol H atoms) / (1.008g H) = 11.10 mol H atoms
O: 88.79g x (1 mol O atoms) / (16.00g O) = 5.549 mol O atoms

3) Change the numbers to whole by dividing by smallest number (which in this case is 5.549)

H= 11.10 mol / 5.549 mol = 2.000
O= 5.549 mol / 5.549 mol = 1.000

Back 19

The simplest ratio of H to O is 2:1

= H20 as the empirical formula

Front 20

Calculate the empirical formula for this substance given the knowns:

56.58% K
8.68% C
34.73% O

Back 20

The simplest ratio of K to C to O is 2:1:3

= K2CO3 as the empirical formulaujh

Front 21

Reactants

Back 21

Substances entering the reaction.

Front 22

Products

Back 22

Substances formed.

Front 23

Balancing Chemical Equations

( Ex: Mercury (II) Oxide ---> mercury + oxygen )

1) Identify the reaction
2) Write the unbalanced equation.
(In this case: HgO ---> Hg + O2)
3) Balance! Hg is balanced because there is 1 Hg atom in each side of the equation but Oxygen isn't. 1 O reactant and 2 O products, it needs to be balanced out.
Select the smallest coefficients that will give the same number of O atoms on EACH side.
A coffiecient placed in the front of the atom multiples EVERY atom in the formula by THAT number)
(So in this case: We place 2 in front of HgO ( 2 HgO)
Now Oxygen is balanced but Hg is not! ( 2 Hg reactants and 1 Hg product)
Put a 2 in front of the Hg product and now the equation and the equation should be balanced.

Back 23

2 HgO ---> 2 Hg + O2

2 Hg atoms |2 Hg atoms
2 O atoms | 2 O atoms

Front 24

Balance the equation:

KCIO3 ----> KCI + O2

Back 24

2 KCIO3 ---> 2 KCI + 3 O2

2 K atoms | 2 K atoms
2 CI atoms |2 Cl atoms
6 O atoms |6 O atoms

Front 25

Balance the equation:

AgNO3 + H2S ----> Ag2S + HNO3

Back 25

2 AgNO3 + H2S ---> Ag2S + 2 HNO3

2 Ag atoms | 2 Ag atoms
2 N atoms | 2 N atoms
6 O atoms | 6 O atoms
2 H atoms | 2 H atoms
1 S atom | 1 S atom

Front 26

Combination Reaction

Back 26

2 reactants combine to give one product.

Front 27

Decomposition Reaction

Back 27

A single substance is decomposed to give 2 or more different substances.

Front 28

Singe Displacement

Back 28

One element reacts with a compound to replace one of the elements.

Front 29

Double Displacement

Back 29

Two compounds exchange partners with each other to produce two different compounds.

Front 30

Exothermic

Back 30

Gives heat

Front 31

Endothermic

Back 31

Absorbs heat

Front 32

Heat of Reaction

Back 32

The quantity of heat produced by a reaction.

Front 33

Hydrocarbons

Back 33

Compounds containing only hydrogen and carbon.

Front 34

Activation Energy

Back 34

The amount of energy that must be supplied to start a chemical reaction.