1. A gaseous element has two isotopes: G-102 with an atomic weight of 102.11 and G-108 with an atomic weight of 108.08. The percent abundance of the heavier isotope is 34.550%. What is the average atomic weight of element G ?
(108.08)(0.34050) = 37.342
+ (102.11)(0.65450) = 66.831
______
average at. wt. = 104.17
2. A new element - Ma - is discovered with three stable isotopes. The average atomic weight is 88.934. The lightest isotope has a mass of 84.889, the middle isotope weighs 86.778 and accounts for 23.70% abundance of the element. The heaviest isotope weighs 90.222 amu. What is the percent abundance of the heaviest isotope?
88.934 = 90.222(1.00 - 0.2370 - X) + 86.778(0.2370) + 84.889(X)
88.934 = 90.222(0.7630 - X) + 86.778(0.2370) + 84.889X
88.934 = 68.84 - 90.222X + 20.57 + 84.889X
-0.476 = -5.333(X)
X = 0.0893 = 8.93% (lighter isotope)
for the heaviest isotope % abundance = 67.37%
3. A sample is analyzed as containing 24.09 grams of potassium, 0.308 moles of manganese,
and 7.42 x 1023 atoms of oxygen. What is the empirical formula?
K Mn O
24.09 g 0.308 moles 7.42 x 1023 atoms
------= 0.616 moles ------= 1.23
39.1 g/mol 6.02 x 1023 atoms/mol
Divide all 3 by 0.308 è 2 : 1 : 4 è K2MnO4
4. A sample of a complex organic protein contains 8.60 x 1022 atoms of carbon, 1.89 x 1023 atoms of hydrogen, 5.16 x 1022 atoms of oxygen, and 2.58 x 1022 atoms of nitrogen. This 4.000-gram sample is known to be 0.007140 moles.
A) What is the empirical formula of this compound?
Hint: don’t worry about the exponent - make all the same power of 10
C H O N
0.860 1.89 0.516 0.258
------= 0.143 ------= 0.314 ------= 0.0857 ------= 0.0429
6.02 6.02 6.02 6.02
Divide all 4 by 0.0429
(3.33 = 3 1/3 7.32 = 7 1/3 2.00 1.00 ) x 3 = 10 : 22 : 6 : 3
C8H22O6N3
B) What is the molecular weight of the empirical formula? 280 g/mole
C) What is the molecular weight of the molecular formula?
4.000 g / 0.007140 moles = 560 g/mol
D) What is the molecular formula? 560 / 256 = 2 è C16H44O12N6
5. B How many sulfur dioxide molecules are there in a 4.00-milliliter sample of this gas under STP conditions? A) 1.49 x 10-22; B) 1.08 x 1020; C) 3.23 x 1020;
D) 3.23 x 1023; E) 1.08 x 1023; F) 5.40 x 1025; G) none of these
0.00400 L 1 mol SO2 6.02 x 1023 molecules
x ------x ------= 1.075 x 1020 molecules
22.4 L gas 1 mol SO2
6. G If a sample of ferric carbonate has 2.25 x 1024 atoms of oxygen in it, what is the mass (in grams) of this sample? A) 8.65; B) 432; C) 865; D) 144; E) 1090; F) 288; G) 121
Fe2(CO3)3
2.25 x 1024 atoms of O 1 mol O 1 mol cpd 291.6 g cpd
x ------x ------x ------
6.02 x 1023 atoms 9 moles O 1 mol cpd
= 121.096 g Fe2(CO3)3
7. 6.06 x 10-23 g What is the mass of a single molecule of hydrogen chloride gas at STP conditions?
. 36.5 g HCl 1 mol HCl
------x ------= 6.06 x 10-23 g per molecule
1 mole 6.02 x 1023 molecules
Consider the following lab information for the first two questions:
Suzy Scientist weighed an empty crucible and cover. She then added some hydrated sample into the crucible, weighed it, and then heated it strongly for several minutes until only the anhydrous form of the salt remained. The crucible and anhydrous salt was then weighed. The identity of the anhydrous salt is known to be cupric carbonate.
mass of empty crucible & cover ……………. 34.6650 grams
= 5.3576 g hydrate
mass of crucible, cover, & hydrated salt …… 40.0226 grams
= 2.4941 g water
mass of crucible, cover, and anhydrous salt .. 37.5285 grams
= 2.8635 g anh. salt
8. 0.02319 mole How many moles of the anhydrous salt were used in this experiment?
2.8635 g 1 mol salt
x ------= 0.02319 mole salt
123.5 g salt
9. 6 How many moles of water are there in every mole of the hydrated salt;
ie. in the formula CuCO3. X H2O, what is X ?
2.4941 g H2O 1 mol water
x ------= 0.1386 mole water
18.0 g water
0.1386 mole water / 0.02319 mole salt = 5.975 : 1 è 6 : 1
10. ___6__HCl + ___1__Al2O3 à __2___AlCl3 + ___3__H2O
11. ___2__C2H2 + ___5__O2 à ___4__CO2 + ___2__ H2O
12. __3___Hg(NO3)2 + ___2__FeCl3 à ___3__HgCl2 + ___2__ Fe(NO3)3
13. 1 NH4OH + 1 HNO3 à 1 NH4NO3 + 1 H2O
14. 3 Sn + 2 H3PO4 à 3 H2 + 1 Sn3(PO4)2
15. 2 Fe + 3H2SO4 à 3 H2 + 1 Fe2(SO4)3
16.. Joe Student mixes 20.0 ml of a 3.00 M solution of silver nitrate with 15.0 ml of a 2.00 M solution of sodium phosphate. He collects and dries the precipitate formed, then weighs the precipitate and finds that it has a mass of 8.00 grams. Answer the following questions based on this information.
A. Write the balanced equation for this reaction.
3 AgNO3(aq) + Na3PO4(aq) à Ag3PO4(ppt.) + 3 NaNO3(aq)
B. How many moles of each reactant were there?
(0.0200 L)(3.00 M) = 0.0600 mol AgNO3
(0.0150 L)(2.00 M) = 0.0300 mol Na3PO4
C. What is the limiting reagent?
0.0300 mol Na3PO4 (3 mol AgNO3 / 1 mol Na3PO4) = 0.0900 mol AgNO3 needed
only have 0.0600 mol AgNO3 so it is the limiting reagent
D. How many moles of excess reagent were left unreacted?
0.0600 mol L.R. (1 mol E.R. / 3 mol L.R.) = 0.0200 mol E.R. needed
0.0300 mol - 0.0200 mol = 0.0100 mol xcs
E. What is the precipitate? Ag3PO4(ppt.)
F. How many moles of precipitate should have been formed theoretically?
0.0600 mol L.R. (1 mol ppt. / 3 mol L.R.) = 0.0200 mol ppt.
[ = 8.37 grams ppt.]
G. What is the percent yield for this reaction?
8.00
------x 100 = 95.6%
8.37
H. What is the concentration of each of the spectator ions (two of them)?
(0.0300 mol) x 3 0.0600 mol
[Na+1] = ------= 2.57 M [NO3-1] = ------= 1.71 M
0.0350 L 0.0350 L
I. What is the concentration of the excess reagent ion?
[PO4-3] = 0.0100 mol / 0.0350 L = 0.286 M
J. How many more moles of the limiting reagent would need to be added to produce stoichiometric amounts?
from C - would need 0.0300 more moles of the L.R.
17. Josephine Student (Joe's twin sister) runs another experiment. She reacts 4.00 grams of solid copper metal with 10.0 ml of concentrated nitric acid. Concentrated nitric acid is 15.8 M. The three products are cupric nitrate, nitric oxide (NO), and water. She collects the NO gas and finds it has a volume of 805 ml at STP conditions. Answer the following questions based on this information.
A. Write the balanced equation for this reaction.
3 Cu(s) + 8 HNO3(aq) à 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(liq)
B. How many moles of each reactant were there?
4.00 g. Cu / 63.5 g/mol = 0.0630 mol Cu (0.0100 L)(15.8 M) = 0.158 mol acid
C. What is the limiting reagent?
0.158 moles acid (3 mol Cu / 8 mol acid) = 0.0593 moles needed Cu
there are 0.0630 moles of Cu - or an excess - so the acid is the L.R.
D. How many moles of excess reagent were left unreacted?
0.0630 moles - 0.0593 moles = 0.0037 moles xcs
E. How many grams of water could be produced, assuming 100% yield?
0.158 mol LR ( 4 mol water / 8 mol LR)(18.0 g water / 1 mol water) = 1.42 g H2O
F. How many liters of nitric oxide gas should have been formed theoretically?
0.158 mol LR ( 2 mol NO / 8 mol LR)(22.4 L NO / 1 mol NO) = 0.885 L NO
G. What is the percent yield for this reaction?
(805 mL / 885 mL) x 100 = 90.96 = 91.0%
H. How many more moles of the limiting reagent would need to be added to produce stoichiometric amounts?
0.0630 moles Cu (8 moles acid / 3 moles Cu) = 0.168 moles acid needed to react
0.168 mol - 0.158 mol = 0.010 mole needed