Download percent composition and formulas

Problem 3.14
• How many moles of cobalt (Co) atoms are
there in 6.00 X109 (6 billion) Co atoms?
PROBLEM 3.24-F
• CALCULATE THE MOLAR MASS OF: (f) Mg3N2
PROBLEM 3.27
• CALCULATE THE NUMBER OF C, H, AND O
ATOMS IN 1.50 G OF OF GLUCOSE (C6H12O6)
PROBLEM 3.30
• The density of water is 1.00 g/ml at 4 Celsius
degrees. How many water molecules are in
2.56 ml of water at this temperature?
Percent composition of an element in a compound =
n x molar mass of element
molar mass of compound
x 100%
n is the number of moles of the element in 1 mole of
the compound
%C =
%H =
%O =
C2H6O
2 x (12.01 g)
46.07 g
6 x (1.008 g)
46.07 g
1 x (16.00 g)
46.07 g
x 100% = 52.14%
x 100% = 13.13%
x 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
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MASS RELATIONSHIPS: PERCENT
COMPOSITION AND FORMULAS
• What is the percent of Hydrogen in H2O?
TYPES OF FORMULAS
• MOLECULAR (C6H1206)
• EMPIRICAL (CH2O)
Problem 3.39
• Tin (Sn) exists in Earth’s crust as SnO2.
Calculate the percent composition by mass of
Sn and O in SnO2.
• Determine the empirical formula of a
compound that has the following
percent composition by mass:
K
24.75, Mn 34.77, O 40.51 percent.
PROCEDURE
• Find grams of each element. CONVERT
GRAMS TO MOLES
• DIVIDE BY SMALLEST NUMBER OF MOLES
• EXPRESS MOLE RATIO
• Determine the empirical formula of a
compound that has the following
percent composition by mass:
K
24.75, Mn 34.77, O 40.51 percent.
Percent Composition and Empirical Formulas
Determine the empirical formula of a compound that
has the following percent composition by mass:
K
24.75, Mn 34.77, O 40.51 percent.
1 mol K
nK = 24.75 g K x
nMn = 34.77 g Mn x
39.10 g K
1 mol Mn
54.94 g Mn
nO = 40.51 g O x
1 mol O
16.00 g O
= 0.6330 mol K
= 0.6329 mol Mn
= 2.532 mol O
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PROBLEM 3.49 A, B
• WHAT ARE THE EMPIRICAL FORMULAS OF THE
COMPOUNDS WITH THE FOLLOWING
COMPOSITIONS? (a) 2.1 PERCENT H, 65.3
PERCENT O, 32.6 PERCENT S; (b) 20.2
PERCENT Al, 79.8 PERCENT Cl
PROBLEM 3.49
• WHAT IS THE EMPIRICAL FORMULA OF A
COMPOUND CONTAINING (A) 2.1 Percent H;
65.3 Percent O; 32.6 percent S
A process in which one or more substances is changed into one or more new substances is a
chemical reaction
A chemical equation uses chemical symbols to show what happens during a chemical
reaction
reactants
products
3 ways of representing the reaction of H2 with O2 to form H2O
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Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on the
left side and the correct formula(s) for the product(s)
on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of each
element the same on both sides of the equation. Do
not change the subscripts.
2C2H6
NOT
C4H12
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Balancing Chemical Equations
3. Start by balancing those elements that appear in only
one reactant and one product.
C2H6 + O2
2 carbon
on left
C2H6 + O2
6 hydrogen
on left
C2H6 + O2
CO2 + H2O
1 carbon
on right
start with C or H but not O
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
multiply H2O by 3
2CO2 + 3H2O
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Balancing Chemical Equations
4. Balance those elements that appear in two or more
reactants or products.
C2H6 + O2
2 oxygen
on left
C2H6 +
multiply O2 by
2CO2 + 3H2O
O2
2C2H6 + 7O2
7
2
4 oxygen
(2x2)
+ 3 oxygen
(3x1)
2CO2 + 3H2O
7
2
= 7 oxygen
on right
remove fraction
multiply both sides by 2
4CO2 + 6H2O
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Balancing Chemical Equations
5. Check to make sure that you have the same number of
each type of atom on both sides of the equation.
2C2H6 + 7O2
4CO2 + 6H2O
4 C (2 x 2)
4C
12 H (2 x 6)
12 H (6 x 2)
14 O (7 x 2)
14 O (4 x 2 + 6)
Reactants
Products
4C
4C
12 H
12 H
14 O
14 O
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PROBLEM 3.59
• BALANCE THE FOLLOWING EQUATIONS.
C + O2  CO
H2 + Br2  HBr
K + H2O  KOH + H2
H2O2  H2O + O2
S8 + O2  SO2
KOH + H3PO4  K3PO4 + H2O
CH4 + Br2  CBr4 + HBr
Problem 3.59
• BALANCE THE FOLLOWING EQUATIONS
(a) C + O2  CO
(c) H2 + Br2  HBr
(i) Zn + AgCl  ZnCl2 +Ag
(k) NaOH + H2SO4  Na2SO4 + H2O
How to “Read” Chemical Equations
2 Mg + O2
2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
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Chemical Equations:
Amounts of Reactants and Products
1.
Write balanced chemical equation
2.
Convert quantities of known substances into moles
3.
Use coefficients in balanced equation to calculate the number of moles of the
sought quantity
4.
Convert moles of sought quantity into desired units
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Problem 3.66
• Silicon tetrachloride (SiCl4) can be prepared by
heating Si in chlorine gas:
Si(s) + Cl2(g) SiCl4(l)
In one reaction 0.507 mole of SiCl4 is produced.
How many moles of molecular chlorine were
used in the reaction.
Problem 3.68
Consider the combustion of butane (C4H10 ):
2C4H10 + 13(O2)  8CO2 + 10H2O
In a particular reaction, 5.0 moles of C4H10 are
reacted with an excess of O2. Calculate the
moles of CO2 formed.
Problem 3.71
• When Potassium Cyanide (KCN) reacts with
acids, a deadly poisonous gas, hydrogen
cyanide (HCN) is given off. Here is the
equation.
KCN(aq)+ HCl(aq)  KCl(aq) + HCN(g)
If a sample of 0.140 grams of KCN is treated with
an excess of HCl, calculate the amount of HCN
formed in grams.
Problem 3.75
• Limestone (CaCO3) is decomposed by heating
to quicklime (CaO) and carbon dioxide.
Calculate how many grams of quicklime can
be produced from 1.0 kg of limestone.
CaCO3  CaO + CO2
Limiting Reagent:
Reactant used up first in the
reaction.
2NO + O2
2NO2
NO is the limiting reagent
O2 is the excess reagent
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• In one process, 124 g of Al are reacted with
601 g of Fe2O3.
2Al + Fe2O3  Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed
OR
g Fe2O3
124 g Al x
mol Fe2O3
1 mol Al
27.0 g Al
mol Al needed
x
Start with 124 g Al
1 mol Fe2O3
2 mol Al
160. g Fe2O3
x
1 mol Fe2O3
g Al needed
=
367 g Fe2O3
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
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Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
2Al + Fe2O3
124 g Al x
1 mol Al
27.0 g Al
x
1 mol Al2O3
2 mol Al
g Al2O3
Al2O3 + 2Fe
102. g Al2O3
x
1 mol Al2O3
=
234 g Al2O3
At this point, all the Al is consumed and Fe2O3
remains in excess.
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PROBLEM 3.83
• Nitric Oxide (NO) reacts with oxygen gas to
form nitrogen dioxide (NO2), a dark brown
gas:
2NO(g) + O2(g)  2NO2(g)
In one experiment, 0.886 moles of NO are mixed
with 0.503 moles of O2(g). Calculate which of the
two is the limiting reagent. Calculate the moles of
NO2 produced.
Reaction Yield
Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
Actual Yield
% Yield =
x 100%
Theoretical Yield
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Percent Yield
• In an organic chemistry lab, 12.4 grams of an
organic compound was produced.
According to calculations, the reaction should
have produced 17.0 grams.
Calculate the percent yield.
Chemistry In Action: Chemical Fertilizers
Plants need: N, P, K, Ca, S, & Mg
3H2 (g) + N2 (g)
2NH3 (g)
NH3 (aq) + HNO3 (aq)
NH4NO3 (aq)
fluorapatite
2Ca5(PO4)3F (s) + 7H2SO4 (aq)
3Ca(H2PO4)2 (aq) + 7CaSO4 (aq) + 2HF (g)
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