Honors Chemistry I
Mole Concept / Molarity / Empirical & Molecular Formulas
Practice Test
Make sure to use the proper number of sig. figs and include the proper units with all answers.
1.____5.664 x 10-23 g _______________________ What is the mass, in grams, of a single molecule of
hydrosulfuric acid?
1 molec. H2S x (1 mol/6.022 x 1023) x (34.0808 g/1 mol) =
2.____107 g_____________ A 30.0-liter sample of dinitrogen pentoxide contains how many grams of oxygen?
30.0 L x (1 mol/22.4 L) x (5 mol O/1 mol N2O5) x (15.9998 g /1 mol) =
3.____5.09 x 1020 molecules _____________
How many molecules of water are present in 80.0
milligrams of ferric chlorate tetrahydrate?
.0800 g x (1 mol/378.2542 g) x (4 mol water/1 mol ferric chlorate tetrahydrate) x (6.022 x 1023/1 mol) =
4._____68.3%____________ What is the percentage of metal in a 2.000-gram sample of barium dichromate?
(241.32/353.32) x 100 =
5._____16.87 g____________
What is the mass of exactly 13.50 liters of carbon monoxide gas at STP
conditions?
13.50 x (1 mol/22.41 L) x (28.010 g/1 mol)=
6.___6.40 x 1023 atoms ______________ What is the number of hydrogen atoms in 34.7 grams of
phosphoric acid?
34.7 g x (1 mol/97.9937 g) x (3 mol H/1 mol phosphoric acid) x (6.022 x 1023/1 mol) =
-27.___13.5 mol oxygen_____________
If you had 3.00 moles of auric carbonate, how many moles of pure
oxygen gas could you get from this sample? [Hint: remember that oxygen gas is
diatomic!]
3.00 mol auric carbonate x (9 mol oxygen/1 mol auric carbonate) x (1 mol O2/2 mol O) =
8.____3.891 x 1019 atoms _______________
sodium tetraborate?
How many total atoms are in exactly 1.000 milligram of
.001000 g x (1 mol/201.217 g) x (13 mol atoms/1 mol sodium tetraborate) x (6.022 x 1023/1 mol)
9.___.892 mol _______________ How many moles of perchloric acid do you need in order to release 10.0
liters of hydrogen gas (H2) at STP?
10.0 L x (1 mol/22.41L) x (2 mol perchloric acid/1 mol hyrdrogen)
10.___2.79 x1023 ions__________________ If a compound of stannic chloride contains 45.53% tin by weight,
how many chloride ions are there in 30.0 grams of this compound?
54.47% chloride
.5447 x 30.0 = 16.341 g Cl x (1 mol/35.453 g) x (6.022 x 1023/1 mol) =
11._____7.53 x 1023 ions ____________________ How many ions (total of both cations and anions) are
there in 0.0250 moles of chromium(III) oxalate?
.0250 mol Cr(III)ox x (5 mol ions/1 mol cmp) x (6.022 x 1023/1 mol) =
12._____1.01 mol___ How many moles of metal can be recovered from 125 grams of strontium molybdate?
SrMoO4
125 g x (1 mol/247.556 g) x (2 mol Metal/1 mol cmp) =
-313.____904 mL_____________
If 300.0 milliliters of a 6.15 M solution of nitric acid is diluted until the
new concentration is only 2.04 M, what will be the final volume of this diluted
solution?
(6.15)(300.0) = (2.04)(V)
14.___Ca3(PO4)2 ___________
What is the empirical formula of a substance that is analyzed and found
to contain 38.77% calcium, 19.98% phosphorus, and 41.25% oxygen?
38.77 g Ca x (1 mol/40.078) = 0.9674/.6451 = 1.500 x 2 = 3
19.98 g P x (1 mol/30.974) = 0.6451/.6451 = 1 x 2 = 2
41.25 g O x (1 mol/15.999) = 2.578/.6451 =3.996 x 2 = 8
Ca3P2O8
15.__119 mL_______________
How much water must be added to 45.0 ml of a 2.00 M solution of
phosphoric acid to produce a 0.550 M solution of this acid?
(2.00)(45.0) = (0.550)(V)
V = 164 mL (new solution volume)
164-45.0 = 119 mL water to be added
16.___N4O8 ______________
4.72 grams of a gas at STP conditions occupies 575 ml. This gas is known
to contain 30.435% nitrogen and 69.565% oxygen. What is the true, or molecular
formula?
.575 L x (1 mol/22.41 L) = 0.0257 mol
MM = 4.72 g/0.0257 mol = 184 g/mol
30.435 g N x (1 mol/14.007 g) = 2.1728 mol/2.1728 = 1
69.565 g O x (1 mol/15.999 g) = 4.3481 mol/2.1728 = 2.0011
NO2 = 46 g/mol
184/46 = 4
17.____2.43 g_____________
How many grams of barium hydroxide need to be added to 750.0 ml of
water in order to produce a 0.0189 M solution of this base?
0.0189 M = x mol/.7500
x = 0.0142 mol x (171.3438 g/1 mol) =
18.____Ba(NO3)2·7H2O ________________When 44.466 grams of a hydrate of barium nitrate is heated,
30.000 grams of the anhydrous residue remains. What is the correct formula of
this hydrate?
30.000 g Ba(NO3)2 x (1 mol/261.338 g) = .11479 mol/.11479 = 1
14.466 g H2O x (1 mol/18.0148 g) = .80301 mol/.11479 = 6.9954
-419.___0.726 M____________
What is the molarity of a solution if 1.35 grams of lithium chlorite is
added to enough water to make 25.0 ml of solution?
1.35 g x (1 mol/74.392) = 0.0181 mol
M = 0.0181/.0250
20.____6.45 L____________ How many liters of water must be added to 4.75 liters of a 10.25 M solution of
potassium chloride in order to end up with 11.20 liters of a 4.347 M solution?
11.20-4.75 =
21.__7.50 M__________________ What is the [S-2] if 2.00 moles of ferric sulfide is added to enough water to
produce 800.0 ml of solution?
Fe2S3
M = 2.00/.8000 = 2.50 M = [Fe2S3]
2.50 x 3 =
22.___Mn2O3 ________________ Oxygen gas is passed over a 5.000-gram sample of solid manganese and
allow to react with this metal. If 7.186 grams of an oxide form of manganese is
produced, what is the empirical formula of this compound?
5.000 g Mn x (1 mol/54.938 g) = 0.09101mol/.09101 = 1x 2 = 2
2.186 g O x (1 mol/15.999 g) = 0.1366 mol/.09101 = 1.501 x 2 = 3
23.____C20H27O6________________
An organic compound is known to contain 48.00 grams of carbon,
24
3.25 x 10 atoms of hydrogen, and 1.20 moles of oxygen atoms. If 4.000 grams
of this compound is known to be 0.01102 moles, what its true molecular formula?
48.00 g x (1 mol/12.011 g) = 3.996 mol/1.20 = 3.33 x 6 = 20
3.25 x 1024 atoms x (1 mol/6.022 x 1023) = 5.40 mol/1.20 = 4.5 x 6 = 27
1.20 mol O/1.20 = 1.00 x 6 = 6
C20H27O6 = 363.4273
MM = 4.000/0.01102 = 363.0