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Chapter 5: Moles and Equations Review {Chemistry}
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Which quantity is equivalent to 39 grams of LiF?
1.
1.0 mole
2.
2.0 mole
3.
0.50 mole
4.
1.5 mole
The gram formula mass is the sum of the gram atomic masses of the atoms that make up a chemical formula. For lithium fluoride this works out to be 7g + 19g = 26g. Hence the mass of 1 mole of LiF is 26g; and the mass of 1.5 moles of LiF must be 1.5 × 26g = 39g.
What is the total number of moles of atoms in one mole of (NH4)2SO4?
1.
10
2.
11
3.
14
4.
15
The compound in question is ammonium sulfate. From the chemical formula we see that it contains 15 atoms per molecule (2 N, 2×4 = 8 H, 1 S, and 4 O). Thus, one mole of the compound contains 15 moles of atoms.
If an equation is balanced properly, both sides of the equation must have the same number of
atoms
Which process represents a chemical change?
1.
melting of ice
2.
corrosion of copper
3.
evaporation of water
4.
crystallization of sugar
The corrosion of copper is an oxidation reaction and is a chemical change. The other processes are just physical changes
Formula Mass
Sum of all masses of all components in a chemical
-If the chemical is simply an atom, it may be referred to as the ATOMIC MASS
-All formula masses are derived from the average atomic mass for each element provided in the periodic table.
-Unit is in u, not g!!
EXAMPLE: The atomic mass of iron is 55.8 u
Gram-Formula Mass
Gram formula mass is most commonly called molar mass. Its abbreviated gfm. to create the gram formula mass, you take a formula mass and change the units from u to g.
EXAMPLE: gfm of carbon dioxide:
(1 x mass of C) + (2 x mass of O) = (1 x 12.0 u) + (2 x 16.0 u) = 44.0 u, which is equal to 44.0 g.
The Mole
1 mol is equal to 6.02 x 10^23 things.
EXAMPLE: if considering water, 6.02 x 10^23 molecules= 1 mole=18.0 g of water (gram formula mass)
Hydrates
A hydrate is an ionic compound that traps water inside of the crystal lattice framework in the spaces between the ions, the water can be removed, leading to an anhydrous salt (a.k.a. anhydrate)
EXAMPLE: CoCl2 * 6H20
("*" means there are two substances)
find gfm:
Co= 58.9 x 1
Cl= 35.5 x 2
H= 1.0 x 12
O= 16.0 x 6
Which all equals 237.9 g
Percent Composition
percentage of the relative mass that element or component of a substance that contributes to the mass of the compound.
% comp. by mass = Mass of part/Mass of whole x 100
Molecular formula
shows the number and kind of atoms present in a substance
-also called the "chemical formula"
-most common formula used
EXAMPLE: C2H4
Structural Formula
shows the number and kind of atoms AND the spatial arrangement of atoms in a substance
-provides the most information about a molecule
-Lewis Dot structures are one common type of structural formula
Empirical Formula
gives the kinds of atoms in the lowest possible ratio in a substance
-always used for ionic compounds
-along with % comp, it can be used to calculate the molecular formula
EXAMPLE: CH2
Determining the Empirical Formula from % comp.
Unknown liquid analyzed, shows 60.0 % C, 13.4% H, and 26.6 % O by mass. Calculate the empirical formula
1. Given Info
-% C= 60.0
-%H= 13.4
-% O=26.6
2. Unknown C? H? O?
3.Equations and Extra Info: Assume you have a 100.0 g sample, change percent to grams
4.Substitute and Solve
60.0 g C x 1 mol C/12.0g C=5.00 mol C
13.4 g H x 1 mol/1.01g H=13.3 mol H
26.6 g O x 1 mol/16.0g O=1.66 mol O
5. Subscripts have to be whole numbers, so divide each # of moles by the smallest subscript
***You can round if you are two tenths above or below a whole #
Determining a molecular formula from an Empirical Formula and GFM
Empirical Formula: P2O5 GFM: 284 g/mol
1. Givens
-empirical formula = P2O5
-Gram Formula Mass=284 g/mol
2.Unknowns
-molecular formula=?
3. Equations and Extra info
-The molecular formula is always a multiple of the empirical formula
-Find the GFM of the empirical formula using the periodic table compare it to the molecular formula's GFM
4.Substitute and Solve
-Find GFM mass of the empirical formula, P2O5
-solve for n which is the number of empirical formulas within the molecular formulas (the ratio between them)
n = empirical formula GFM/molecular GFM
142.0g/284.0 g, will give 0.5:1, but after multiplying by 2, it gives 1:2 ratio
5. Multiply the empirical formula by this ratio, you get P4O10
Stoichiometry
is the study of the relationships between substances in a balanced chemical reaction
EXAMPLE: Always start with a balanced equation
_N2 + _3H2 yields _2NH3
Form a conversion factor using mole ratios
If 3.5 moles of N2 are used, how much NH3 is produced?
3.5 mol N2 x 2 mol NH3/1 mol N2 = 7.0 mol NH3
Balancing Equations
All changes in chemistry including chemical reactions must obey the Law of Conservation Mass, Energy and Charge must be balanced
Balancing Equations by Inspection Process & Considerations:
1. Count the number of atoms of each element on each side of the arrow
2. Change the coefficient to make changes in the number of atoms to reach equality between the amount of reactant and product for each element
3. Reduce the coefficient ratio to the lowest possible, as needed
TIPS FOR STEP 2:
Keep polyatomic ions together if they stay together throughout the reaction
If you need a .5 or 1/2 of a coefficient you must multiply all coefficients by 2 to get whole numbers at the end of balancing
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