Chemistry Chapter 3 Test Questions Flashcard
(Figure 3.3). A chemist determines from the amounts of elements that 0.0654
mol ZnI2 can form. How many grams of zinc iodide is this?
prepared by a precipitation reaction (Figure 3.4). In a preparation, 45.6 g
of lead(II) chromate is obtained as a precipitate. How many moles of PbCrO4
H 6.73 percent
O 53.3 percent
the manufacture of plastics (Figure 3.5), and a water solution of the compound is used to
preserve biological specimens. Calculate the mass percentages of the elements in formaldehyde
(give answers to three significant figures).
composition obtained in the previous example (40.0% C, 6.73% H, 53.3% O).
the Greek khroma, meaning “color”; see Figure 3.9.) Sodium dichromate is the
most important commercial chromium compound, from which many other
chromium compounds are manufactured. It is a bright orange, crystalline substance.
An analysis of sodium dichromate gives the following mass percentages:
17.5% Na, 39.7% Cr, and 42.8% O. What is the empirical formula of this compound?
(Sodium dichromate is ionic, so it has no molecular formula.)
H, and 53.4% O. Determine the empirical formula. The molecular mass of acetic acid was
determined by experiment to be 60.0 amu. What is its molecular formula?
substance from which the metal can be profitably obtained.) The free metal
is obtained by reacting hematite with carbon monoxide, CO, in a blast furnace.
Carbon monoxide is formed in the furnace by partial combustion of carbon. The
Fe2O3(s) + 3CO(g) —> 2Fe(s) + 3CO2(g)
How many grams of iron can be produced from 1.00 kg Fe2O3?
Formerly chlorine was produced by heating hydrochloric acid with pyrolusite (manganese
dioxide, MnO2), a common manganese ore. Small amounts of chlorine may
be prepared in the laboratory by the same reaction (see Figure 3.13):
4HCl(aq) + MnO2(s) —-> 2H2O(l) + MnCl2(aq) + Cl2(g)
How many grams of HCl react with 5.00 g of manganese dioxide, according to this
containing manganese(II) acetate (catalyst) under pressure at 60C.
2CH3CHO(l) + O2(g) —-> 2HC2H3O2(l)
In a laboratory test of this reaction, 20.0 g CH3CHO and 10.0 g O2 were put into a reaction
vessel. a. How many grams of acetic acid can be produced by this reaction from these amounts
of reactants? b. How many grams of the excess reactant remain after the reaction is complete?
1. In sodium fluoride, there is one atom of sodium and one atom of fluorine. The molar mass will then be:
(1 atom x 23 grams/mole of sodium) + (1 atom x 19 grams/mole of fluorine) = 42 grams/mole of sodium fluoride
2. In potassium hydroxide, there is one atom of potassium, one atom of hydrogen, and one atom of oxygen. The molar mass will then be (1 x 39 grams) + (1 x 1 gram) + (1 x 16 grams) = 56 grams/mole of potassium hydroxide
3. In copper (I) chloride, there is one atom of copper and one atom of chlorine. The molar mass is then (1 x 63.5 grams) + (1 x 35.5 grams) = 99 grams/mole of copper (I) chloride
4. In manganese (IV) oxide, there is one atom of manganese and two atoms of oxygen. The molar mass is then (1 x 55 grams) + (2 x 16 grams) = 87 grams/mole of manganese (IV) oxide
5. In calcium sulfate, there is one atom of calcium, one atom of sulfur, and four atoms of oxygen. The molar mass is then (1 x 40 grams) + (1 x 32 grams) + (4 x 16 grams) = 136 grams/mole of calcium sulfate
6. In magnesium phosphate, there are three atoms of magnesium, two atoms of phosphorus, and eight atoms of oxygen. (The formula is Mg3(PO4)2). The molar mass is then (3 x 24 grams) + (2 x 31 grams) + (8 x 16 grams) = 262 grams/mole of magnesium phosphate
1. sodium fluoride
2. potassium hydroxide
3. copper (I) chloride
4. manganese (IV) oxide
5. calcium sulfate
6. magnesium phosphate
FW = 12.01 amu + (4 x 1.008 amu) + 16.00 amu = 32.042 = 32.0 amu
b. FW of NO3 = AW of N + 3(AW of O)
= 14.01 amu + (3 x 16.00 amu) = 62.01 = 62.0 amu
c. FW of K2CO3 = 2(AW of K) + AW of C + 3(AW of O)
= (2 x 39.10 amu) + 12.01 amu + (3 x 16.00 amu)
= 138.210 = 138 amu
d. FW of Ni3(PO4)2 = 3(AW of Ni) + 2(AW of P) + 8(AW of O)
= (3 x 58.70 amu) + (2 x 30.97 amu) + (8 x 16.00 amu)
= 366.040 = 366 amu
Find the formula masses of the following substances to
three significant figures.
a. methanol, CH3OH
b. nitrogen trioxide, NO3
c. potassium carbonate, K2CO3
d. nickel phosphate, Ni3(PO4)2
FW of NH4NO3 = 2(AW of N) + 4(AW of H) + 3(AW of O)
= (2 x 14.01 amu) + (4 x 1.008 amu) + (3 x 16.00 amu)
= 80.052 amu
The molar mass of NH4NO3 = 80.05 g/mol.
Ammonium nitrate, NH4NO3, is used as a nitrogen
fertilizer and in explosives. What is the molar mass of
1.16 x 10^-2 mol H2O
Calcium sulfate, CaSO4, is a white, crystalline powder.
Gypsum is a mineral, or natural substance, that is a hydrate of
calcium sulfate. A 1.000-g sample of gypsum contains 0.791 g
CaSO4. How many moles of CaSO4 are there in this sample? Assuming
that the rest of the sample is water, how many moles of
H2O are there in the sample? Show that the result is consistent
with the formula CaSO42H2O.
( sample of mass/sample in C of mass) x 100 = 86.27%
A 1.836-g sample of coal contains 1.584 g C. Calculate
the mass percentage of C in the coal.
A fertilizer is advertised as containing 14.0% nitrogen (by
mass). How much nitrogen is there in 4.15 kg of fertilizer?
Percent C = (C of mass/CO of mass) x = 42.9%
Percent O = 100.000% – 42.878%C = 57.122 = 57.1%
Calculate the percentage composition for each of the following compounds (three significant figures).
Which contains more carbon, 6.01 g of glucose, c6h12o6, or 5.85 g of ethanol, C2H6O?
Mol Os = 2.16 g Os x 1 mol Os/190.2 g Os =0.01136 mol (smaller number)
Mol O =0.0456 mol
Integer for Os = 0.01136 ÷ 0.01136 = 1.000
Integer for O = 0.0456 ÷ 0.01136 = 4.01
Within experimental error, the empirical formula is OsO4
An oxide of osmium (symbol Os) is a pale yellow solid. If 2.89 g of the compound contains 2.16 g of osmium, what is its empirical formula?
Mol Mn =0.5078 mol (smallest number)÷ 0.5078 =1
Mol O =2.031 mol÷ 0.5078 =4
Potassium manganate is a dark green, crystalline substance
whose composition is 39.6% K, 27.9% Mn, and 32.5% O,
by mass. What is its empirical formula?
Formula weight = 44.08 amu
88.5 amu ÷ 44.1 amu = 2.006, or 2
Therefore, the molecular formula is (C2H6N)2, or c4h12n2.
Putrescine, a substance produced by decaying animals,
has the empirical formula C2H6N. Several determinations of molecular
mass give values in the range of 87 to 90 amu. Find the
molecular formula of putrescine
1.2 mol CO2
Butane, C4H10, burns with the oxygen in air to give carbon
dioxide and water.
2C4H10(g) + 13O2(g) –> 8CO2(g) + 10H2O(g)
What is the amount (in moles) of carbon dioxide produced from
0.30 mol C4H10?
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l ) —> 4KOH(s) + 3O2(g)
If a reaction vessel contains 0.25 mol KO2 and 0.15 mol H2O, what is the limiting reactant? How many moles of oxygen can be produced?
5. formula unit