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Day 263. Express the composition of each of the following compounds as the mass percents
of its elements (just percent composition)
a. Formaldehyde, CH2O
b. Glucose, C6H12O6
c. acetic acid, HC2H3O2
64. Considering your answer to Exercise 63, which type of formula, empirical or
molecular, can be obtained from elemental analysis that gives percent composition.
65. Give the empirical formula for each of the following compounds: Empirical is the
lowest one- you can not divide it any further- like math!- Molecular is a multiple of
empirical
a. C6H8O6
b. C6H6
c. C2H2
d. P4O10
e. C6H12O6
f.
HC2H3O2
70. A Sample of Urea contains 1.121 g N, 0.161 g H, 0.480 g C, and 0.640 g O. What is
the empirical formula of urea ( Remember this is where you will use the AMU on the
periodic table to convert to moles, then you divide by the lowest value to set it to
one, then divide the others by that number to find the ratios between the other ones:
[ 1.121 g N ( 1 mole of N/ 14.00 g) = 0.008 moles N] { Carbon mole is 0.004 moles
C- so there are 2 N for every 1 C})
72. A compound containing only sulfur and nitrogen is 69.6% S, by mass; the molar
mass is 184 g/mol. What are the empirical and molecular formulas of the
compound: So remember that % can be replaced with g- same steps as above
except this time get the molecular mass of the empirical formula. Then take the
molecular mass given in the problem and divide the empirical mass into it- that
multiple is how many times you multiply the formula by in order to get the correct
formula)
75. Many homes in rural America are heated by propane gas, a compound that contains
only carbon and hydrogen. Complete combustion of a sample of propane produced
2.641 g of Carbon dioxide and 1.442 g of water as the only products. Find the
empirical formula of propane. Remember that a combustion reaction is one with a
CH (hydrocarbon) reacting with Oxygen- Carbon dioxide and water are produced.
To start this one find the amount of carbon in the carbon dioxide- percent
composition! Take that value and multiple by the given amount of mass. That is the
amount of Carbon in the Carbon dioxide that was given off. Repeat with the
Hydrogen. Then use these values to convert to moles like in the previous problems.
Divide by the the smallest mole value to get the ratios. Then find the empirical
formula- On this one DO NOT ROUND! Find a multiple! Maybe THREE!)
78. A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68
mg of the compound yields 16.01 mg carbon dioxide and 4.37 mg of water. The
molar mass of compound is 176.1 g/mol. What are the empirical and molecular
formulas of the compound. Convert to grams first- milli means 1000 then
Remember that a combustion reaction is one with a CH (hydrocarbon) reacting with
Oxygen- Carbon dioxide and water are produced. To start this one find the amount
of carbon in the carbon dioxide- percent composition! Take that value and multiple
by the given amount of mass. That is the amount of Carbon in the Carbon dioxide
that was given off. Repeat with the Hydrogen. Then use these values to convert to
moles like in the previous problems. Divide by the the smallest mole value to get
the ratios. Then find the empirical formula- On this one DO NOT ROUND! Find a
multiple! Maybe THREE!) you also have to get to the molecular formula by doing the
molecular mass!
79. Write a balanced equation that describes each of the following Donʼt forget your
seven diatomics! and remember to balance Charges!
a. Iron metal reacts with oxygen to form rust [iron (II) oxide]
b. Calcium metal reacts with water to produce aqueous calcium hydroxide and
hydrogen gas
c. Aqueous barium hydroxide reacts with aqueous sulfuric acid to produce solid
barium sulfate and water
80. Give the balanced equation for each of the following chemical reactions Donʼt forget
your seven diatomics! Remember to balance Charges!
a. Glucose ( C6H12O6) reacts with oxygen gas to produce gaseous carbon dioxide
and water
b. Solid iron (II) sulfide reacts with gaseous hydrogen chloride to form solid iron (III)
chloride and hydrogen sulfide
81. Balance the following equations 1. Metals first, 2. Non metals next, 3. Hydrogen and
Oxygen Last
a. Cu + AgNO3 --> Ag + Cu(NO3)2
b. Zn + HCl
--> ZnCl + H2
c. Au2S3 + H2 --> Au + H2S
82. Balance the following equations 1. Metals first, 2. Non metals next, 3. Hydrogen and
Oxygen Last
a. Ca(OH)2 + H3PO4 --> H2O + Ca3(PO4)2
b. Al(OH)3 + HCl
--> AlCl3 + H2O
c. AgNO3 + H2SO4 --> Ag2SO4 + HNO3
83. Balance the following equations representing combustion reactions 1. Metals first, 2.
Non metals next, 3. Hydrogen and Oxygen Last- sometimes you have to double to
get to the desired results- no 1/2 solutions
a. C12H22O11 + O2 --> CO2 + H2O
b. C6H6 + O2
c. Fe + O2
--> CO2 + H2O
--> Fe2O3
d. C4H10 + O2 --> CO2 + H2O
e. FeO + O2
--> Fe2O3
84. Balance the following equations 1. Metals first, 2. Non metals next, 3. Hydrogen and
Oxygen Last
a. Cr + S8 --> Cr2S3
b. NaHCO3 --> Na2CO3 + CO2 + H2O
c. KClO3
--> KCl + O2
d. Eu + HF --> EuF3 + H2
90. Over the years, the thermite reaction has been used for welding railroad rails, in
incendiary bombs, and to ignite solid fuel rocket motors. The reactions is
Fe2O3 + 2 Al --> 2Fe + Al2O3
What masses of iron (II) oxide and aluminum must be used to produce 15.0 g
iron? What is the maximum mass of aluminum oxide could be produced.
Balance the equation first, Then convert grams given in the problem to moles (15.0 g Fe
( 1 mole Fe/ 55.6 g Fe)-- then convert that one mole to the desired mole - use the
coefficents from the balanced equation for this one- Then convert the moles to grams of
the desired quantity- Reverse of the first step (x g of desired/ 1 mole of desired)----- you
have three sets of these to do in order to find the amount)
Remember monday you get more questions and also on Wednesday it will be individual
ANSWERS
63. C- 40%, H- 6.67%, O-53.5% Notice that all of them have the same percentages even though they have various molecular
formulas! Huge idea there!!
64. a. Empirical
b. Molecular c. molecular
65. a. C3H4O3, b. CH, c. CH, d. P2O5, e. CH2O, f. CH2O- Remember that empirical is the simplest formula, Molecular formula is
a multiple of it or can be reduced.
70.
CN2OH- does not matter which order they are written in.....
71. did not have to do
72. SN is the empirical and the Molecular formula is S4N4
73. did not have to do
74. Did not have to do
75.
C3H8
76. Did not have to do
77. Did not have to do
78.
C3H4 empirical and the molecular is C12H16
79. A. 4Fe + 3O2 --> 2Fe2O3, B. Ca + H2O --> Ca(OH)2 + H2, C. Ba(OH)2 + H2SO4 --> BaSO4 + 2H2O
80. A. C6H12O6 +6O2 --> 6CO2 + 6H2), b. Fe2S3 + 6 HCl --> 2 FeCl3 + 3 H2S
81.
82.
83.
Balance the following equations
a.
Cu + 2AgNO3 --> 2Ag + Cu(NO3)2
b.
Zn + 2HCl
c.
Au2S3 + 3H2 --> 2Au + 3H2S
--> ZnCl + H2
Balance the following equations
a.
3Ca(OH)2 + 2H3PO4 --> 6H2O + Ca3(PO4)2
b.
Al(OH)3 + 3HCl
c.
2AgNO3 + H2SO4 --> Ag2SO4 + 2HNO3
--> AlCl3 +3 H2O
Balance the following equations representing combustion reactions
a.
2C12H22O11 + 24O2 --> 24CO2 + 22H2O
b.
2C6H6 + 15O2
c.
4Fe + 3O2
d.
2C4H10 + 13O2 --> 8CO2 + 10H2O
e.
4FeO + O2
--> 12CO2 + 6H2O
--> 2Fe2O3
--> 2Fe2O3
84.
Balance the following equations
a.
16Cr + 3S8 --> 8Cr2S3
b.
2NaHCO3 --> Na2CO3 + CO2 + H2O
c.
2KClO3
d.
2Eu + 6HF --> 2EuF3 + 3H2
--> 2KCl + 3O2
90. 21.47 grams of Fe2O3, 7.27 g of Aluminum, 13.75 grams of Al2O3:
conservation!
Remember that products equal reactants based on law of