Download Study Buddy Unit II- Constant Velocity

Study Buddy: Determining Formulas & Nomenclature
Terms
Description
Percent Composition
Representation
The percent by mass of each element present in a
compound.
Empirical Formula
The simplest ratio of atoms present in a compound
Molecular Formula
The actual number of atoms of each element that occur
in one molecule of the compound
Law of Definite
Proportions
Law of Multiple
Proportions
A chemical compound always contains the same
proportion of elements by mass regardless of amount of
sample
Whenever two elements form more than one
compound, the elements will combine in whole number
ratios and the compounds will have different properties
Mass of element
X 100
Mass of compound
Molecular
C6H12O6
H2 O 2
CH4
Empirical
CH2O
HO
CH4
Water is always 88.8%
O and 11.2% H by
mass
CO or CO2 exist, but not
CO1.5
Representations:
1. Draw ionic models for the ions when they react, name them and write the formula:
a. Magnesium and chlorine
MgCl2
Mg+2
Magnesium chloride
Cl-1
b. Iron(III) and oxygen
Fe2O3
Fe+3
O-2
Iron(III) oxide
Li2SO4
c. Lithium and sulfate
Li+1
SO4-2
Lithium sulfate
2. Explain when each of these would be used in chemical names:
a. Roman numerals
transition metals with more than one valency
b. Prefixes
covalent compounds (no mono- on 1st element)
c. –ide ending
the anion of ionic cmpds and the 2nd name of covalent cmpds
d. Ending unchanged
the cation of ionic cmpds, the 1st name of covalent cmpds & polyatomics
3. Write the name or formula for each of the following compounds:
Name
Carbon tetrachloride
Tin(IV) oxide
Dinitrogen tetroxide
Iron(II) phosphate
Phosphorus pentafluoride
Methane
Sodium nitrate
Calcium sulfide
12/8/2015
Ionic or Covalent
Covalent
Ionic
Covalent
Ionic
Covalent
Covalent
Ionic
Ionic
Formula
CCl4
SnO2
N2O4
Fe3(PO4)2
PF5
CH4
NaNO3
CaS
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4. What is the name and formula of the compound that forms when aluminum reacts with the sulfate
ion? Aluminum sulfate, Al2(SO4)3
a. How many atoms of each element are in one formula unit of this compound?
Al _2__
S _3__ O _12__
b. How many moles of each element are in one mole of this compound?
Al _2__
S _3__ O _12__
c. What is the percent composition by mass for each of the elements in this compound?
Molar mass of Al2(SO4)3 = 2(27.0)+3(32.1)+12(16.0) = 342.3 g/mol
% Al = 2(27.0)/342.3 x 100 = 15.8% Al
% S = 3(32.1)/342.3 x 100 = 28.1% S
% O = 12(16.0)/342.3 x 100 = 56.1% O
5. Lead (II) chromate, PbCrO4, was used as a pigment in paints. How many moles of the compound
are in 75.0 g of lead (II) chromate?
Molar mass of PbCrO4 = 323 g/mol
X mol PbCrO4
75.0 g PbCrO4
=
1 mol PbCrO4
323 g PbCrO4
0.232 moles PbCrO4
a. How many atoms of oxygen are present in the 75.0 g of PbCrO4?
X atoms oxygen
(4)(6.02x1023) atoms oxygen
5.59x1023 atoms oxygen
=
1 mol PbCrO4
0.232 mol PbCrO4
6. What is the difference between a molecular formula and an empirical formula?
A molecular formula shows the actual number of atoms in one molecule of a compound, whereas
an empirical formula shows the simplest ratio of atoms in the compound. The molecular formula
represents a compound, but the empirical may not represent an actual compound that exists.
7. Describe how a chemist would use an empirical formula to determine the molecular formula?
Once the chemist has determined empirical formula by using the percent composition of each of the
elements, the chemist would use the actual molar mass of the compound to determine the
“multiplier” by comparing it to the molar mass of the empirical formula. The chemist would then
multiply each of the subscripts by the multiplier.
8. A compound was analyzed and found to contain 22.2% nitrogen, 1.6% hydrogen, and 76.2%
oxygen. What is the empirical formula of the compound?
Nitrogen – 22.2% x 100g = 22.2g / 14.0 g/mol = 1.59 mol 1.59 mol/1.59 mol = 1
HNO3
Hydrogen – 1.6% x 100g = 1.6g /1.0 g/mol =
1.6 mol 1.6 mol/1.59 mol = 1
Oxygen – 76.2% x 100 =
76.2g / 16.0 g/mol = 4.76 mol 4.76 mol/1.59 mol = 3
9. A compound composed of hydrogen and oxygen is found to contain 0.59 g of hydrogen and 9.44 g
of oxygen. The molar mass of this compound is 34.0 g/mol. Find the empirical and molecular
formulas.
Hydrogen – 0.59g / 1.0 g/mol = 0.59 mol 0.59/0.59 = 1 empirical formula  HO
Oxygen – 9.44g / 16.0 g/mol = 0.590 mol 0.590/0.59 = 1
Molar mass of HO  1.00 + 16.0= 17.0 g/mol
Molar mass of compound  34.0 g/mol
12/8/2015
Multiplier = 34.0/17.0 = 2
Empirical Formula  H2O2
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