Bell Work, Dec 1 – Dec 5, 2015
Molar Mass, Moles, Calculating Moles
Bell Work, Monday, Nov 30, 2015
1. The density of oxygen gas at O°C and 1 atm pressure is 1.43
g/liter, whereas the density of hydrogen gas under these
conditions is 0.089 g/liter. How many times more massive is
one molecule of oxygen than one molecule of hydrogen?
Determine the relative masses
𝒈
𝟏.𝟒𝟑
𝟏𝑳
𝒈
𝟎.𝟎𝟖𝟗
𝟏𝑳
= 16 time massive
H = 1, O = 16
2. If 12.5 g of hydrogen gas combine with 100 g of oxygen to make
112.5 g of water, what is the relative mass of oxygen to hydrogen?
𝟏𝟎𝟎𝒈 𝑶
=8gO:1gH
𝟏𝟐.𝟓 𝒈 𝑯
3. Why is the mass of oxygen in #2 half the value in question 1?
Because the hydrogen in question 2 is H2 not H1. So the mass of
𝟏𝟎𝟎𝒈 𝑶
H1 is ½ of 12.5 = 6.25
= 16 g O : 1 g H
𝟔.𝟐𝟓 𝒈 𝑯
Bell Work, Monday, Nov 30, 2015
4. What is molar mass for elements.
• It is the relative mass in grams of one standard lump of
elements. A standard lump of an element has 1 mole of atoms.
• Scientist choose carbon to compare relative masses.
• By definition, something called the carbon-12 isotope (a carbon
atom with 6 protons and six neutrons) has molar mass of
exactly 12 grams.
5. What is the relative mass and the molar mass of an element
that is ½ the mass of C-12?
½ of 12 = 6, so the relative mass = 6, the molar mass = 6 grams
6. What is the relative mass of an element that is 2 x the mass of C-12?
2 x 12 = 24, so the relative mass = 24, the molar mass = 24 grams
7. What is a mole?
The number of particles in a mole = 6.022 x 1023
The molar mass of any element contains 6.022 x 1023 atoms
the molar mass in grams of any element or compound = 1 mole
Bell Work, Tuesday, Dec 1, 2015
1. Find the number of nitrogen, carbon, hydrogen, &
oxygen atoms represented by the formula (NH4)2CO3
(NH4)2CO3 = 2(NH4) and 1(CO3)
2(NH4): 2•N1 =
2• 1N = 2 N,
2•H4 =
(2 • 4H) = 8 H
Plus 1 CO3 = 1 atom of carbon and 3 atoms of oxygen
2 nitrogens, ____
8 hydrogens, ____
1 carbons and ____
3
____
oxygen atoms.
Bell Work, Tuesday, Dec 1, 2015
1. Find the number of nitrogen, carbon, hydrogen, & oxygen
atoms represented by the formula (NH4)2CO3
8 H, ____
1 C and ____
3 O
____
2 N , ____
2. Now find the molar mass of each element on the periodic table.
g Molar mass of H =1.008
g
Molar mass of N = 14.01
________
______
g
g Molar mass of O = 16.00
Molar mass of C = 12.01
________
______
3. The molar mass of (NH4)2CO3 is found multiplying the molar
mass from the periodic table by the number of each element in the
formula and adding up the masses.
For a
molecule, the
Molar mass of two N’s is 2 x 14.01 = 28.02 g
formula
8.064 g
mass and
Molar mass of 8 H’s is 8 x 1.008 =
the molar
12.01
g
Molar mass of one C is 1 x 12.01 =
are the same
Molar mass of three O’s is 3 x 16.0 = 48.00 g
96.094 g
Bell Work, Wednesday, Dec 2, 2015
1. A 4.07 g sample of NaI contains how many moles of NaI?
_____ x
= ______
(given) times factor = (answer)
Grams  moles
Molar mass of Na = 22.99 grams, Molar mass of I = 126.9
Molar mass of NaI = 149.9,
1 mole of NaI = 149.9 g/ mol NaI
mol NaI
4.07 g NaI x 1 mol NaI = ?______
149.9 NaI
= 0.027 mol NaI
2. A 4.07 g sample of NaI contains how many moles (mol) of Na?
1 molecule of NaI = 1 Na and 1 I, I mol of NaI = 1 mol Na, 1 mol I
4.07 g of NaI = 0.027 mol NaI, 0.027 mol NaI = 0.027 mol of Na, 0.027 mol I
4.07 g NaI contains 0.027 moles of Na
Bell Work, Wednesday, Dec 2, 2015
3. A 4.07 g sample of NaI contains how many atoms of Na?
Grams  moles  # of atoms
Molar mass of Na = 22.99 grams, Molar mass of I = 126.9
Molar mass of NaI = 149.9, 1 mole of NaI = 149.9 g NaI
1 mole of NaI = 6.022 x 1023 molecules of NaI
1 mole of Na = 6.022 x 1023 atoms Na
6.022 x 1023 atoms Na
0.027 mol Na x
= ? atoms Na
1 mol Na
= 1.63 x 1022 atoms Na
Bell Work, Thursday, Dec 3, 2015
1. Use Avogadro’s number to find how many atoms are
there in 0.00150 moles Zn?
Avogadro’s number = 6.022 x 1023
1 mole of Zn = 6.022 x 1023 atoms of Zn
23 atoms Zn
6.022
x
10
atoms Zn
0.00150 mol Zn x
= ? ______
1 mol Zn
= 9.03 x 1020 atoms Zn
(given)
times conversion factor =
(answer)
Bell Work, Thursday, Dec 4, 2015
2. What is the mass of 100 million atoms of gold? Could you
mass this on a balance?
To little to weigh out.
#of atoms  moles  Grams
Avogadro’s number = 6.022 x
1 mole of Au = 6.022 x 1023 atoms of Au
1 mol Au
8
1.00
x 10 atoms Au x
_____________
23
1 mole of Au =
196.97 g
1023
x 10 -16 mol Au
= 1.66
______
6.022 x 10 atoms Au
(given) times conversion factor = (answer)
196.97 g Au
-16
-14
1.66
x
10
mol
Au
3.27
x
10
_____________
x
= ______
grams Au
1 mol Au
1.00  108 atoms Au 
1 mol
 197.0 g = 3.27 x 10-14 g Au
6.02 x 1023 atoms
1 mol
Bell Work, Thursday, Dec 4, 2015
This bell work is for bonus points & is for review
3. The relative mass of an atom can be found by comparing the
mass of the atom to the mass of
a.
one atom of carbon-12. c.
a proton.
b.
one atom of hydrogen-1. d.
uranium-235.
carbon atoms
4. A mole is the number equal to the number of _______
in exactly 12 g of pure __________
carbon-12.
5. The carbon-12 atom is assigned a relative mass of exactly
a.
1 g.
c.
12 g.
b.
6 g.
d.
100 g.