Chemistry Chapter 7 Answers

Flashcard maker : Patrick Turner
indicates the relative # of atoms of each kind in a chemical compound
Chemical Formula
The Simplest ratio of cations and anions
Formula Unit
Distribute: Al2(SO4)3
Al2 S3 O12
Monatomic Ions
ions formed from a single atom
Group 1A Na (balance charge)
+1 charge Na+1
Group 4A C (balance charge)
+0 because it is even distance from noble gasses
Group 5A N (balance charge)
-3 charge C-3
Cation always occurs _______ an anion in a chemical equation
before
compounds composed of 2 elements
binary compounds
In a binary compound the total number of positive and negative charges must be ______
equal
Write the formula for Aluminum Oxide
1. Al O
2. Al+3 O-2
3. cross charge numbers Al2 O3 (and drop charge)
Rules in naming binary ionic compounds (3)
1. name the cation first and anion second, cation is always first in formula
2. Monatomic cations use the element name
3. monatomic anions take their name from the root of the element plus -ide as a suffix
Write the name of K2O
Potassium Oxide (K O2) switch numbers and drop sign
Write the name of CuCl2
Copper (II) Chloride
(Cu Cl2) work backwards Cu+2 Cl-1
use chart given in class to find charge
write the formula and name of Fe+4 O-2
Iron (IV) Oxide
(Fe O2) Oxygen can never exist as a -1 charge and always -2 so double the cation to balance the doubling of oxygen
ions that are made up of more than one atom
polyatomic ions
the charge given to a polyatomic ion refers to _______________
the whole group of atoms
Write the formula for Ammonium Chloride
NH4Cl
NH+1 Cl-1
switch numbers and drop charge signs
NH4= ammonium
Write the name of K2CrO4
Potassium Chromate
K+1 (CrO4)-2 charge
switch numbers and drop charge signs
CrO4= Chromate
____________ of naming molecular compounds is based on the use of prefixes (type of system)
old system
Rules for the prefix system of nomenclature of binary molecular compounds are (3)
1. The first element is given a prefix only if it contributes more than one atom to a molecule of the compound ex: K2 O4 = Dipotassium
2. the second element is named by combining a prefix based on the number of atoms contributed, the root name of the element, and ending in -ide
3. The (o or a) at the end of a prefix is usually dropped when the word following the prefix begins with a vowel
ex: O4= Tetroxide not Tetraoxide
Numeral Prefixes 1-10 (name them)
Mono- Di- Tri- Tetra- Penta- Hexa- Hepta- Octa- Nona- Deca-
Write the name of As2O5
Diarsenic Pentoxide
As2 O5
Name the two categories most acids are classified under
Binary acids and Oxyacids
Acids that consist of 2 elements, usually hydrogen and one of the halogens
binary acids
acids that contain hydrogen, oxygen, and a third element (usually a non metal)
oxyacids
Name the acid HF
hydrofluoric acid
Name the acid HCl
hydrochloric acid
Name the acid HBr
hydrobromic acid
Name the acid HI
hydroiodic acid
Name the acid H3PO4
phosphoric acid
Name the acid HNO2
nitrous acid
Name the acid hno3
nitric acid
Name the acid H2SO3
sulfurous acid
Name the acid H2SO4
sulfuric acid
Name the acid CH3COOH
acetic acid
Name the acid HClO
hypochlorous acid
Name the acid HClO2
chlorous acid
Name the acid HClO3
chloric acid
Name the acid HClO4
perchloric acid
Name the acid H2CO3
carbonic acid
the mass of any unit represented by a chemical formula, whether the unit is a molecule, a formula unit, or an ion is known as __________
formula mass (AMU) atomic mass unit
formula unit or ion is the sum of the average atomic mass of all the atoms in its formula
formula mass of any molecule
Find the Formula mass of Potassium Chlorate
Potassium=K
Chlorate=ClO3

K- 1 * 39 =39
Cl-1 * 35=35
O-3 * 16=48
39+35+48=122
122 AMU
the 1st column is the element, 2nd is the amount of atoms of that element, 3rd is the atomic mass of those given elements, 4th is the result of the number of atoms multiplied by the atomic mass. Then add all the solutions together to get atomic mass

What is the molar mass of Barium Nitrate (g/mol)
Barium=Ba
Nitrate=NO3
since oxygen is diatomic it must be even therefor the formula is Ba(NO3)2
Ba-1 * 137=137
N- 2 * 14=28
O- 6 * 16=96
137+28+96=261
261 g/mol
the _____________ of a compound can be used as a conversion factor to relate an amount in moles to a mass in grams of a given substance
molar mass
What is the mass, in grams, of 2.50 mol of Oxygen gas? (O2)
grams > mole > molecules
2.50 mol O2 / 1 32 grams O2 / 1 mol O2
2.50*32=80 grams
sigifgs (3) 80.0 grams
32 came from the atomic mass of oxygen, it was doubled because there were 2 atoms of oxygen in the equation, and the number 2.50 was given
A: Find the molar mass of Ibuprofen (C13H18O2) B: find out how many moles of ibuprofen are in a 33gram bottle. C: How many molecules are in the bottle? D: What is the total mass, in grams, of carbon in 33 grams of ibuprofen? (use 206.31 as molar mass after step A)
A: Ibuprofen = C13H18O2
C- 13 * 12= 156
H- 18 * 1 = 18
O- 2 * 16 = 32
156+18+32=(206)
for the purpose of the example we are using 206.31 as molar mass because it is 100% accurate and not a rounded representation of the atomic mass of the other elements
B: 33grams ibuprofen / 1 1mol ibuprofen / 206.31grams ibuprofen 33grams / 206.31 grams = (0.16 mols) ibuprofen
C: 0.16 mol ibuprofen / 1 6.02*10^23 molecules ibu. / 1mol ibu. 0.16* 6.02*10^23= (9.6*10^22 molecules of ibuprofen)
D: 0.16mol ibuprofen / 1 13 mol C / 1 mol ibuprofen 12 g C / 1 mol C 12*13*0.16=24.96=25(sigifg)
(25grams C)
mass percentage formula
mass of element in compound / mass of entire compound *100= percent of element in compound
the percentage by mass of each element in a compound is known as the _______
percentage composition
Find the percentage composition of Copper (I) Sulfide, Cu2S.
Copper= Cu2
Sulfur = S
Copper- 2 *64=128
Sulfur- 1 * 32=32
128+32=160
128 / 160= 0.8=80% Copper
32 / 160= 0.2=20% Sulfur
the ______________ consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole number ratio of the elements in a compound
empirical formula
Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula for this compound
32.38g Na / 23 = 1.408mol Na / 0.7078 = 2
22.65g S / 32 = 0.7078mol S / 0.7078 = 1
44.99g O / 16 =2.812mol O / 0.7078 = 4
first # in the column was the percent composition, second was the atomic mass of the element, after dividing the 2 next the answer was divided by the smallest number (0.7078) to the nearest whole number
Na2SO4- Sodium sulfate
the actual formula of the molecular compound
molecular formula
molecular formula (formula)
x (empirical formula) = molecular formula
the empirical formula of a compound of phosphorous and oxygen was found to be P2O5. experimentation shows that the molar mass of this compound is 283.89 g/mol What is the molecular formula of this compound?
step 1: find mass of empirical formula
step 2: divide molar mass of molecular formula by mass of empirical formula
step 3: the answer of step 2 is X in the equation x(empirical formula)= molecular formula
step 4: utilize formula
1: P- 2 * 31=62
O- 5 * 16=80
62+80=142
2: 283.89 / 142 = 2 (always a whole number)
3: x=2
4: 2(P2O5)= molecular formula
(P4O10)

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