Answers additional questions 7.2
1.
What is the smallest part of water that has all the chemical characteristics of water?
One molecule of water. Water is a covalent compound
2.
What is the ratio of the mass of 1 carbon atom to the mass of 1 hydrogen atom?
1 carbon atom
1 hydrogen atom
=
12.0 amu
1.0 amu
What units are used to express atomic mass?
atomic mass units
What is the ratio of the mass of 1 mole of hydrogen atoms to 1 mole of carbon atoms?
1 mole carbon
1 mole hydrogen
=
12.0 grams
1.0 gram
What units are used to express the mass of a mole?
grams
Why does the ratio of hydrogen mass to carbon mass stay the same for atoms and moles?
Because you have an equal number of particles of carbon and hydrogen the mass ratio remains the same as the
individual particles.
How many individual particles are in a mole?
1 mole always contains 6.02 x 1023 particles.
3.
How many moles of iron metal are in 16.8 grams of iron?
1 mole Fe = 55.8 grams
16.8 g * 1 mole Fe / 55.8 g = 0.301 moles Fe
4.
How many grams of magnesium are in 0.55 moles of magnesium?
1 mole Mg = 24.3 g
0.55 moles Mg * 24.3 g / 1 mole Mg = 13.4 g Mg
5.
How many moles of aluminum, sulfur, and oxygen are in 1 mole of aluminum sulfate?
Al2 (SO4 )3
2 moles Al
3 moles S
12 moles O
6.
Find the percentage by mass of calcium, carbon, and oxygen in calcium carbonate.
CaCO3
1 Ca * 40.1 = 40.1
1 C * 12.0 = 12.0
3 O * 16.0 =48.0
40.1 / 100.1 = 40%
12.0 / 100.1 = 12%
48.0 / 100.1 = 48%
Total mass = 100.1
How many grams of carbon are in 275 grams of calcium carbonate?
275 g CaCO3 * 12.0 g C / 100.1 g CaCO3 = 33.0 g C
7.
How many grams of calcium hydroxide will it take to contain 25.5 grams of calcium?
Ca(OH)2
1 Ca * 40.1
2 O * 16.0
2 H *1.0
= 40.1
=32.0
=2.0
Total = 74.1
25.5 g Ca * 74.1 g Ca(OH)2 / 40.1 g Ca =
8.
47.1 g Ca(OH)2
How many moles of aluminum are in 275 grams of aluminum nitrate?
Al(NO3 )3
1 Al * 27.0 = 27.0
3 N * 14.0 = 42.0
9 O * 16.0 =144.0
Total = 213.0 g
275g Al(NO3 )3 * 1 mole Al(NO3 )3 /213.0g * 1 mole Al / 1 mole Al(NO3 )3
= 1.29 mole Al
9.
Find the empirical formula of the compound with 2.4 grams carbon, 0.6 grams hydrogen, and 1.6 grams
oxygen.
2.4 g C / 12.0 amu = 0.2
0.2 / 0.1 = 2
0.6 g H / 1.0 amu = 0.6
0.6 / 0.1 =6
1.6 g O / 16.0 amu = 0.1
0.1 / 0.1 = 1
empirical formula is C2 H6 O1
10.
Early rockets used hydrazine, 93.3% nitrogen and 6.7% hydrogen by mass as fuel.
Find the empirical formula of hydrazine
93.3 % N / 14.0 amu = 6.66
6.66 / 6.66 = 1
6.7 % H / 1.0 amu = 6.7
6.7 / 6.66 = 1
empirical formula = N1 H1
If the molecular mass of hydrazine is 30 amu find the molecular formula.
mass of empirical formula = 1 N * 14.0 amu + 1 H * 1.0amu = 15.0 amu
molecular mass = 30.0 amu
empirical mass = 15.0 amu
molecular formula = N2 H2
= 2