weds march 5 , 2014
DESCRIPTION
Weds March 5 , 2014. Due: HW 7C, Lab Reports Today: Determining Chemical Formulas Empirical Formulas Molecular Formulas Friday: Magnesium Oxide Lab Bring your own goggles if you don’t want to wear the ones in the classroom set! . Necessary skills: . Multiplication Division - PowerPoint PPT PresentationTRANSCRIPT
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• Due: HW 7C, Lab Reports• Today: Determining Chemical Formulas• Empirical Formulas• Molecular Formulas
• Friday: Magnesium Oxide Lab • Bring your own goggles if you don’t want to wear the ones
in the classroom set!
Weds March 5, 2014
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•Multiplication•Division•Convert mass to moles
Necessary skills:
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• A Molecular formula includes the symbol of elements and the number of atoms of each element in that molecule
• Empirical formula includes symbols of elements in compounds with subscripts that show the smallest possible whole numbers that describe the atomic ratio
Calculation of Chemical Formulae
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• Ionic compounds – formula unit IS ALREADY the smallest whole-number ratio
• Molecular compounds (covalent bonds) – molecule is not always the smallest whole number ratio!
Determining Chemical Formula
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•Molecular formula: CH4
•Empirical formula: •CH4
Methane
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•Molecular formula: C2H6
•Empirical formula: •CH3
Ethane
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•Molecular formula: H2O
•Empirical formula: •H2O
Water
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•Molecular formula: C6H6
•Empirical formula: •CH
Benzene
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• 1. Determine Mass Composition• If given percentages: Use percentage composition to convert to a mass
composition (assume 100 g sample so percent is equal to the mass in g)• If given mass: Skip this step
• 2. Convert mass to moles for each element• 3. Find the smallest whole-number ratio by dividing each number of
moles by the smallest number• 4. These whole numbers are the subscripts in your compound
Calculation of Empirical Formula
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• A compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula.• 1. Use % composition to get to mass composition• 32.38% Na = 32.38 g Na
• 22.65% S = 22.65 g S
• 44.99% O = 44.99 g O
Example 1 – Step 1
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• A compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula.• 2. Convert mass to moles• 32.38 g Na / (22.989 g/mol) = 1.408 mol Na
• 22.65 g S / (32.065 g/mol) = 0.706 mol S smallest • 44.99 g O / (15.999 g/mol) = 2.812 mol O
Example 1 – Step 2
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• A compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula.• 3. Find smallest ratio by diving by smallest number of moles• Na: 1.408 mol / 0.706 = 1.99 = 2• S: 0.706 mol / 0.706 = 1• O: 2.812 mol / 0.706 = 3.98 = 4• 4. Write empirical formula• Empirical formula = Na2SO4
Example 1 – Step 3
These numbers become the subscripts in the empirical formula.
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• A compound contains 4.43 g phosphorus and 5.72 g oxygen. Find the empirical formula.• 1. Use % composition to get to mass composition (you
are already there!)
• 4.43 g P
• 5.72 g O
Example 2 – Step 1
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• A compound contains 4.43 g phosphorus and 5.72 g oxygen. Find the empirical formula.• 2. Convert mass to moles
• 4.43 g P / (30.974 g/mol) = 0.143 mol P smallest
• 5.72 g O / (15.999 g/mol) = 0.358 mol O
Example 2 – Step 2
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• A compound contains 4.43 g phosphorus and 5.72 g oxygen. Find the empirical formula.• 3. Find smallest ratio by diving by smallest number of
moles
• P: 0.143 mol / 0.143 = 1 x 2 = 2
• O: 0.358 mol O / 0.143 = 2.5 x 2 = 5• THEY MUST BE WHOLE NUBMERS! YOU CAN’T HAVE
HALF AN ATOM! Multiply by 2 (or 3, if the decimal is .333 or .666) to get whole numbers
Example 2 – Step 3
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ROUNDING HELPS• If the number ends in .98, .99, .01, or .02 Round to nearest whole
number• If the number ends in .50, .51, .52, .49, .48, etc Round to .5 then
multiply by 2• If the number ends in .33, .32, .34, etc Round to .3 then multiply by
3• If the number ends in .66, .65, .64, etc Round to .6 then multiply by
3
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• A compound contains 4.43 g phosphorus and 5.72 g oxygen. Find the empirical formula.• 4. Write empirical formula
• P = 2
• O = 5
• Empirical formula: P2O5
Example 2 – Step 4
These numbers become the subscripts in the empirical formula.
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• If you know the empirical formula and the molecular mass of a compound, you can determine the molecular formula.
• There are four pieces of information. If you know three, you can solve for the fourth. • 1. Empirical formula• 2. Molecular formula• 3. Empirical mass (is the mass of the empirical formula)• 4. Molecular mass (is the mass of the molecular formula)
Calculation of Molecular Formula
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Molecular Formula
Molecular Mass
Empirical Formula
Empirical Mass
Relationship between
Masses (x)Methane CH4 CH4
Ethane C2H6 CH3
Water H2O H2O
Benzene C6H6 CH
Calculation of Molecular Formula
16.05
30.08
18.02
78.12
16.05
15.04
18.02
13.01
1
2
1
6
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The relationship between masses (x) will tell you the relationship between formulas. Find x by:
x = molecular mass empirical mass
What is x?
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• 1. Calculate the empirical mass (like you are used to doing.)
• 2. Solve for x using molecular mass (given in the problem) and empirical mass.
• 3. Multiply the empirical formula by x x(empirical formula) = molecular formula
Steps
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• Determine the molecular formula of the compound with an empirical formula of CH and a molecular mass of 78.110 amu. • Empirical formula= CH• Molecular formula= ??• Empirical mass= mass C + mass H = 12.01 + 1.01 = 13.02 amu• Molecular mass= 78.110 amu
• x = molecular mass = 78.110 = 5.999 formula mass 13.02• 6(CH) = C6H6
Example 1
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• A sample of a compound with a molar mass of 34.00 g/mol consists of 0.44 g H and 6.93 g O. Find its molecular formula.
• Empirical formula: Must determine from data given in the problem (See p.6 of notes) • Molecular formula: ??• Empirical mass: Can determine from empirical formula• Molecular mass: 34.00
• H: 0.44 g H x (I mol H) = 0.436/0.433 = 1 (1.01 g H) HO (empirical formula) • O: 6.93 g O x (I mol O) = 0.433/0.433 = 1 (16.00 g O)
• x = molecular mass = 34.00 = 1.99 empirical mass HO (16.00 + 1.01)
• 2(HO) =H2O2
Example 2