CP Chemistry Midterm Study guide
Name:_______________________________
Part I: Vocab (define the following terms)
Matter
Anything that has mass and takes up space
Compound
2 or more elements chemically combined in a
fixed ratio
Element
A pure substance that cannot be broken down
by physical or chemical means
Density
The amount of mass per unit volume (physical
property)
Heterogeneous mixture
a combination of substances that are not
chemically combined and that does not have
uniform composition
Homogeneous mixture
A combination of substances that are not
chemically combined but has uniform
composition throughout
Filtration a technique that uses a porous
barrier to separate a solid from a liquid
Chromatography
A technique that is used to separate the
components of mixture based on the
components’ ability to be drawn across the
surface of another material
Distillation
A technique that can be used to physically
separate most homogeneous mixtures based
on the differences in the boiling points of the
substances
Chemical property / change
The ability of a substance to react with another
substance
Diatomic gas (O2,N2,F2,I2,Cl2,Br2)
A molecule that contains two identical atoms
Synthesis reaction
A chemical reaction in which two or more
substances form one product
Physical property / change A type of change
that does not alter the substance’s chemical
composition
Atomic number
Number of protons in the nucleus of an atom
Mass number
Number of protons and neutrons in an atom
Isotope
Atoms of the same element that have different
numbers of neutrons
Percent error
The ratio of an error to an accepted value
Noble gas
An extremely unreactive group 18 element
Halogen
A highly reactive group 17 element
Metal
An element that is solid at room temp, a good
conductor of heat and electricity, are shiny,
ductile and malleable
Metalloid
An element that has physical and chemical
properties of both metals and nonmetals
Ductile
Capable of being drawn out into a wire
Malleable
Able to be hammered into a sheet
Ion
An atom or group of atoms with a positive or
negative charge
Polyatomic ion
An ion made up of two or more atoms bonded
together
Oxyanion
A polyatomic ion that contains oxygen
Binary compound
A compound consisting of two elements
Decomposition reaction A single compound
breaks down into two or more substances
Combustion reaction
An exothermic chemical reaction that occurs
when a substance reacts with oxygen
Single replacement reaction A chemical
reaction in which the atoms of one element
replace the atoms of another element in a
compound
Double replacement reaction – a chemical
reaction in which the ions of two substances
are exchanged and produces a solid, gas or
water
meniscus
the curve of an upper surface of a liquid:
measurement taken from the bottom (or top) of
curve
Part II: Describe the following:
1.
Atomic theory
Dalton’s: matter is composed of extremely small particles called atoms
Atoms are indestructible and indivisible
Atoms of a given element are identical
Atoms of one element are different from those of another element
Different atoms combine in simple, whole-number ratios to form compounds
In chemical reaction, atoms are separated, combined or rearranged
Modern:
Atoms are indestructible by chemical means
Isotopes exist (atoms of the same element can differ in their number of neutrons)
2. Atomic structure (use protons, neutrons, electrons)
Spherically shaped, with a small, dense nucleus of positively charged protons and neutral
neutrons surrounded by one or more negatively charged electrons
3. Law of conservation of mass
Matter cannot be created nor destroyed during a chemical reaction
4. The difference between an observation, hypothesis and conclusion
An observation is information gained through the senses (qualitative or quantitative)
A hypothesis is an educated guess that has not been tested.
A conclusion is a judgement based on information obtained.(has been tested)
Part III: Know the following formulas / information
1. Density = grams / volume
2. Percent error = error
X 100%
Accepted value
3. When adding or subtracting, exponents must be the same. (10x)
4. When multiplying, exponents are added (103 X 105 = 108)
5. When dividing, exponents are subtracted (105 ÷ 103 = 102)
6. When multiplying or dividing, the answer should have the same number of significant figures as
the value with the least number of significant figures.
7. When adding or subtracting: 5.06 + 4.3 = 9.3 (cannot go past the tenths place because the value
4.3 does not go past the tenths place.
8. a. How to draw a best fit line.
b. regression equation: y = mx + b (y = dependent variable, x = independent variable, m= slope,
b = y intercept)
9. When measuring, one estimated value is allowed.
Part IV: Solve the following problems
1. A 5mL sample of water has a mass of 5g. What is the density of water? 5g÷5ml = 1 g/ml
2. A cube of aluminum with volume = 5 cm3 has a mass of 20g. What is the density of
aluminum?20g÷5cm3 = 4g/cm3
3. What is the volume of a sample that has a mass of 20g and density of 4 g/mL? 4g/ml ÷ 20g=
0.2ml = 2 X10-1 ml
4. Express the following in scientific notation:
a. 900 = 9.0 X 102
b. 6,500,000 = 6.5 X106
c. 0.00000086 = 8.6 X10-7
5. How many significant figures in the following numbers?
a. 304.0 = 4
b. 0.0067 = 2
c. 604,000 = 3
d. 4.00500 X 108= 6
e. 0.03005 = 4
6. 3 X 102 + 4 X 103 = 3 X 102 + 40 X 102 = 43 X 102 = 4.3 X 103
7. 5 X 10-4 - 7 X 10-2 = 0.05 X 10-2 – 7 X 10-2 = -6.95 X 10-2 = 7 X 10-2
8. 3 X 107 X 9 X 103 = 27 X 1010 = 3 X 1011
9. 4.5 X 10-4 ÷ 9.0 X 103 = 0.5 X 10-7 = 5 X 10-8
10. Three measurements are taken: 34.5m, 38.4m, 35.3m. The actual measurement is 36.7m.
What are the percent errors for the measurements taken?
34.5 = (36.7 – 34.5) ÷ 36.7 X 100% = 6.00%
38.4 = (38.4 – 36.7) ÷ 36.7 X 100% = 4.63%
35.3 = (36.7 – 35.3) ÷ 36.7 X 100% = 3.82%
11. 5g of tin react with hydrochloric acid, producing 8.1g of tin chloride and hydrogen gas. How
much hydrochloric acid reacted? 8.1g – 5g = 3.1 g of HCl
12. Compound I contains 5.63g tin and 3.37 g chlorine. Compound II contains 2.5g tin and 2.98g
chlorine. Are the compounds the same?
Compound I percent composition:
5.63 + 3.37 = 9 g total
Tin = 5.63 ÷ 9 = 0.625 (100%) = 60%
Chlorine = 3.37 ÷ 9 = 0.374 (100%) = 37.4% = 40%
Compound II percent composition:
2.5 + 2.98 = 5.48 g total = 5.5g total
Tin = 2.5 ÷ 5.5 = 0.46 (100%) = 46%
Chlorine = 2.98 ÷ 5.5 = 0.54 (100%) = 54%
They are not the same compound.
13. Silver has two naturally occurring isotopes. Ag-107 has a relative abundance of 51.82% and a
mass of 106.9 amu. Ag-109 has a relative abundance of 48.18% and a mass of 108.9 amu.
Calculate the atomic mass of silver.
Ag – 107 = (51.82%)(106.9amu) = 55.40
Ag-109 = (48.18%)(108.9amu) = 52.50
Atomic mass of silver = 55.40 + 52.50 = 107.9
14. What is the atomic number for the following elements?
a. at atom with 37 electrons = 37
b. an atom with 72 protons = 72
c. an atom with 1 electron = 1
d. an atom with 6 neutrons and a mass number of 12 = 12-6 = 6
15. Predict the result of the following reactions using the activity series:
a. Al + FeCl3
AlCl3 + Fe
b. Br2 + 2LiI
LiBr + I2
c. Cu + MgSO4
no reaction (copper is not reactive enough to replace magnesium – it
is lower than magnesium on the activity series)
16. How many electrons do the following ions have:
a. Mg+2 12 – 2 = 10
b. Na+1 11-1=10
c. O-2 8+2 = 10
d. N-3 7 + 3 = 10
17. What are the products of the following double replacement reactions?
a. LiI + AgNO3
LiNO3 + AgI
b. BaCl2 + K2CO3
BaCO3 + 2KCl
18. Calculate the mass in grams of 2.22 mol of Ti.
2.22 (47.87 g/mol) = 106.3 g
19. How many moles of Co are in 7.65g?
7.65g ÷ 1 mol / 58.9g = 0.130 = 1.30 X 10-1 mol
20. What is the molar mass(gram formula mass) of zinc oxide (ZnO)?
65.39g/mol + 16.00g/mol = 81.39g/mol = 8.139 X 101 g/ mol
21. A compound is analyzed and found to have 69.58% Ba, 24.32% O, and 6.09% C. What is the
compound’s empirical formula?
Step 1: convert grams to moles by dividing by mass number on periodic table:
Ba69.58g
O24.32g
137.3g
16.00g
C6.09g
12.01g
Ba0.5mol
O1.52mol
C0.5mol
0.5mol
0.5mol
0.5mol
Step 2: Divide each mole subscript by the smallest value to convert decimals to whole
number ratios.
Ba1
O3
C1
Empirical formula is: BaO3C, or as you would know it by, BaCO3 (barium carbonate).
22. . A compound is found to have 26.76% C, 2.21% H and 71.17% O.
a. What is it’s empirical formula?
Step 1: convert grams to moles by dividing by mass number on periodic table:
C26.76g
H2.21g
12.01g
1.01g
O71.17g
16.01g
C2.22mol
H2.19mol
O4.48mol
2.19mol
2.19mol
2.19mol
Step 2: Divide each mole subscript by the smallest value to convert decimals to whole
number ratios.
C1
Empirical formula:
H1
CHO2
O2
b. It has a molar mass of 90.04 g/mol. What is its molecular formula?
(compare empirical formula mass to the molecular formula mass)
Empirical formula mass: C + H + O2 = 12 + 1 + 2(16) = 45g
Compare: 45g/mol to 90g/mol : 90 is twice 45 so the molecular formula is twice the empirical
formula:
2(CHO2) = C2H2O4
23. What is the empirical formula for glucose, C6H12O6? = CH2O
24. How many atoms are in the following:
a. 1 mol of Zn 6.02 X 1023 = 6 X 1023
b. 3 mol of O2 3 (6.02 X 1023) = 2(18.02 X 1023) = 3.6 X 1024
c. 2 mol of silver 2 (6.02 X 1023) = 12.02 X 1023 = 1 X 1024
25. How many moles are in the following?
a. 40g of gold 40 g ÷ 200 g/ mol = 0.2 mol = 2 X 10-1 mol
b. 5 g of NaCl 5 g ÷ (20 g/mol + 40 g/mol) = 0.08 mol = 8 X 10-2 mol
c. 6.89g of CuSO4 6.89 g (63.6g/mol + 32.1g/mol + (16.0 g/mol) (4)) = 0.043 mol = 4.3 X 10 -2
mol
26. Analysis of a hydrate of Iron (III) chloride revealed that in a 10.00g sample of the hydrate, 6.00g is
anhydrous iron (III) chloride and 4.00g are water. Determine the formula and name of the hydrate.
Step 1: determine the formula for Iron(III) chloride: Fe3+Cl-1 = FeCl3
6.00g FeCl3
162.4g
4.00g H2O
18.01g
Step 2: Convert 6.00g FeCl3 to moles and 4.00g H2O to moles (molecular mass of FeCl3 = 55.9 + 3(35.5)
= 162.4g/mol; molecular mass of H2O = 2(1.01) + 16 = 18.01g/mol)
6.00gFeCl3 ÷ 162.4g/mol = 0.037 mol FeCl3
4.00gH2O ÷ 18.01g/mol = 0.222 mol H2O
0.037mol FeCl3
0.037
0.222 mol H2O
0.037
Step 3: Divide both by the smallest value to get a whole number ratio.
1 FeCl3
6H2O
The name of this hydrate is Iron(III) chloride hexahydrate.
27. Consider the hydrate: CuSO4 5H2O
a. What is the formula mass?
Cu + S + 4(O) + 5(H2O)
63.5g + 32.1g + 4(16.0g) + 5(18.0g) = 249.6g
b. What is the percent water?
Step 1: Calculate the mass of water: 5(18.0g) = 90.0g
Step 2: Divide the mass of water by the total mass of the compound and multiply by 100%:
90.0g ÷ 249.6g = 0.36 X 100% = 36%
c. How many grams of water would be found in 470.0g of this hydrate?
If the formula is 36% water, then any amount of the formula will contain 36% water:
470.0g (36%) = 169.2g H2O There are 169.2g H2O in 470.0g of this hydrate.
d. How many grams of sulfur will combine with 6.5 g of copper to make this hydrate?
Step 1: Examine the formula CuSO4
5H2O
For every 1 atom of Copper there is one atom of Sulfur. They exist in a 1:1 ratio.
Step 2: Convert 6.5g of copper to moles by dividing by its mass number:
6.5g ÷ 63.5g/mol = 0.10mol copper
Step 2: If there is a 1:1 ratio between copper and sulfur in the formula, then for every 0.10mol of
copper in the formula, there will be 0.10mol of sulfur. 1:1 = 0.10:0.10
Step 3: Convert 0.10 mol of sulfur to grams by multiplying by its mass number:
(0.10mol sulfur) ( 32.1g/mol sulfur ) = 3.21g of sulfur
28. Examine the following reaction:
4A +
13 B
5 C + 17 D
If 0.87 moles of A react, fine the number of moles of B, C, and D.
Balanced chemical equations are mole ratios. For every 4 moles of A that react, 13 moles of B will
react, 5 moles of C will be produced and 17 moles of D will be produced. If you know the amount of
any one of these: A,B,C,D, you can determine the amount of all the rest by using the ratios in the
chemical equation.
Given: 0.87 mol A
Want to find: moles of B,C,D (we will calculate them all separately)
B
0.87 mol A
1
X
13 mol B
4 mol A
=
X
5 mol C
4 mol A
=
X
17 mol D
4 mol A
=
2.83 mol B
C
0.87 mol A
1
1.09 mol C
D
0.87 mol A
1
3.70 mol D
29. Balance the following equation:
___2___ NaOH + ___1__ CuCl2
___2___NaCl + _______ Cu(OH)2
a. How many grams of copper (I) chloride will react with 5.67g of NaOH?
Plan: convert to moles, solve using the mole ratios in the chemical equation, convert back to grams
Step 1: 5.67g NaOH to moles (divide by the formula weight of NaOH = 23.0 + 16.0 + 1.01 = 40.0g)
5.67g NaOH ÷ 40.0g/mol = 0.14 mol NaOH
Step 2:
0.14 mol NaOH
X
1 mol CuCl2
= 0.07 mol CuCl2
1
2 mol NaOH
Step 3: Convert 0.07 mol CuCl2 to grams. (multiply by the formula weight of CuCl2 = 63.5 + 2(35.5) =
134.5g)
(0.07 mol of CuCl2)(134.5 g/mol) = 9.415g CuCl2
So, 5.67g of NaOH will react with 9.42g CuCl2.
b. How many grams of NaCl will be produced?
Plan: convert to moles, solve using the mole ratios in the chemical equation, convert back to grams
Step 1: 5.67g NaOH to moles (divide by the formula weight of NaOH = 23.0 + 16.0 + 1.01 = 40.0g)
5.67g NaOH ÷ 40.0g/mol = 0.14 mol NaOH
Step 2:
0.14 mol NaOH
X
2 mol NaCl
= 0.14 mol NaCl
1
2 mol NaOH
Step 3: Convert 0.14 mol NaCl to grams. (multiply by the formula weight of NaCl = 23.0 + 35.5 = 58.5)
(0.14 mol of NaCl)(58.5 g/mol) = 8.19g NaCl
So, for every 5.67g of NaOH that react, 8.19 g of NaCl will be produced.