Chemical Quantities Word Scramble
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Question | Answer |
Avogadro's Number | 6.02 x 10{23} (written with 3 SF) |
The mole | Particles such as atoms, molecules, and ions are counted by. |
1 mole of carbon, 1 mole of aluminum, 1 mole of sulfur all contain | 6.02 x 10{23} atoms |
1 mole of an element | 6.02 x 10{23} atoms of that element |
Conversion factor between the moles of a substance and the number of particles it contains | 6.02 x 10{23} particles and 1 mole 1 mole 6.02 x 10{23} particles |
Example: we use Avogadro's # to convert 4.00 moles of sulfur to atoms of sulfur | 4.00 moles S atoms x 6.02 x 10{23} S atoms = 2.41 x 10{24} S atm 1 mole S atoms |
Example: we use Avogadro's # to convert 3.01 x 10{24} molecules of CO2 to moles of CO2 | 3.01 x 10{24} CO2 molecules x 1 mole CO2 molecules 6.02 x 10{23} CO2 molecules= 5.00 moles of CO2 molecules |
In a molecule of aspirin, chemical formula C9H8O4 | 9 carbon atoms, 8 hydrogen atoms, 4 oxygen atoms |
Subscripts also state the number of moles of each element in 1 mole of aspirin | 9 moles of C atoms, 8 moles of H atoms, 4 moles of O atoms |
Atoms in 1 molecule C9H8O4 | 9 atoms of C, 8 atoms of H, 4 atoms of O |
Moles of each element in 1 mole C9H8O4 | 9 moles of C, 8 moles of H, 4 moles of O |
Conversion factor | 9 moles C or 1 mole C9H8O4 1 mole C9H8O4 9 moles C |
Conversion factor | 8 moles H or 1 mole C9H8O4 1 mole C9H8O4 8 moles H |
Conversion factor | 4 moles O or 1 mole C9H8O4 1 mole C9H8O4 4 moles O |
Molar Mass | The quantity in grams that equals the atomic mass of that element. |
Example: 1 mole of carbon (C) atoms I mole of carbon atoms, we would weigh out 12.01 g of carbon | First find the atomic mass of 12.01 on the periodic table for carbon (C) |
The molar mass of carbon is found by looking at the atomic mass on the periodic table. | Carbon (C) = 12.01 g of carbon from the periodic table atomic mass |
Carbon (C) | 6.02 x 10{23} atoms of C = 1 mole of C atoms = 12.01 g of C atoms |
Molar Mass of a Compound | Multiply the molar mass of each element by its subscript in the formula, and add the results. |
Find the molar mass of Li2CO3 Obtain the molar mass of each element | 6.94 g Li 12.0 g C 16.0 g O 1 mole Li 1 mole C 1 mole O |
Multiply each molar mass by # of moles (subscript) in the formula Grams from 2 moles of Li | 2 moles Li x 6.94 g Li 1 mole Li = 13.8 g of Li |
Grams from 1 mole of C | 1 mole C x 12.0 g C 1 mole C = 12.0 g of C |
Grams from 3 moles of O | 3 moles O x 16.0 g O 1 mole O = 48.0 g of O |
Calculate the molar mass by adding the masses of the elements. Molar mass of Li2CO3 | 13.8 g of Li + 12.0 g of C + 48.0 g of O = 73.8 g |
Calculations Using Molar Mass | 1 mole of Mg = 24.3 g of Mg |
Created by:
anderson76
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