Do Now: If one stick of Juicy Fruit gum weighs 3.0g, what percent of the total mass of the gum is...
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Transcript of Do Now: If one stick of Juicy Fruit gum weighs 3.0g, what percent of the total mass of the gum is...
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Do Now:
If one stick of Juicy Fruit gum weighs 3.0g, what percent of the total mass of the gum is sugar?
![Page 2: Do Now: If one stick of Juicy Fruit gum weighs 3.0g, what percent of the total mass of the gum is sugar?](https://reader036.fdocuments.us/reader036/viewer/2022062516/56649e0d5503460f94af6c09/html5/thumbnails/2.jpg)
Hydrate Lab – for tomorrow
Pre-lab Title Purpose Materials Procedure Data table
Hydrate Anhydrous
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Hydrates
Methane hydrate
Hydrates, like zinc acetate dihydrate, Zn(C2H3O2)2 * 2H2O are commonly found in skin care products such as moisturizer, shampoo and lip balm.
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What is the % H2O in nickel chloride dihydrate, NiCl2 * 2H2O?
Element # g/mol (molar mass)
TOTAL
Ni 1 58.69 g/mol 58.69g
Cl 2 35.45 g/mol 70.90g
H2O 2 18.02 g/mol 36.04g
MOLAR MASS=
165.63g/mol NiCl2 *
2H2O
€
36.04 g H2O
165.63 g NiCl2 * 2H2O
⎛
⎝ ⎜
⎞
⎠ ⎟%H2O = = 21.76% H2O
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% Mg = x 10024.31 g95.21 g
Percentage CompositionPercentage Composition
Mg
magnesium
24.305
12Cl
chlorine
35.453
17
Mg2+ Cl1-
MgCl2
1 Mg @ 24.31 g= 24.31 g2 Cl @ 35.45 g= 70.90 g 95.21 g
25.52% Mg
74.48% Cl
(by mass...not atoms)
It is not 33% Mg and 66% Cl
% = x 100part
whole
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Empirical and Molecular Formulas
A pure compound always consists of the same elements combined in the same proportions by weight.
Therefore, we can express molecular composition as PERCENT BY WEIGHTPERCENT BY WEIGHT.
Ethanol, C2H6O 52.13% C 13.15% H 34.72% O
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Different Types of Formulas Molecular Formula – shows the real # of atoms in one molecule
or formula unit
Empirical Formula – shows smallest whole number mole ratio
**Sometimes the empirical &molecular formula can be the same
Structural Formula- molecular formula info PLUS bonding electron and atomic arrangement
C6H6
CH
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Calculating Empirical formula
Percent to mass (assume 100g) Mass to mole (molar mass) Divide by the smallest (ratio)
Multiply til’ whole*
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Empirical FormulaQuantitative analysis shows that a compound contains 50.04% carbon, 5.59% hydrogen, and 44.37% oxygen.
Find the empirical formula of this compound.
= 4.17 mol C
= 5.59 mol H
= 2.77 mol O
/ 2.77 mol
/ 2.77 mol
/ 2.77 mol
= 1.5 C
= 2 H
= 1 O
C3H4O2
50.04% C
5.59% H
44.37% O
50.04g C
5.59g H
44.37g O
€
1 mol C
12.01 g C
⎛
⎝ ⎜
⎞
⎠ ⎟
€
1 mol H
1.01 g H
⎛
⎝ ⎜
⎞
⎠ ⎟
€
1 mol O
16.00 g O
⎛
⎝ ⎜
⎞
⎠ ⎟
Step 1) % g Step 2) g mol Step 3) mol mol
Step 4) multiply til whole
*2
*2
*2
X
X
X
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Empirical FormulaQuantitative analysis shows that a compound contains 66.75% copper, 10.84% phosphorusand 22.41% oxygen.
Find the empirical formula of this compound.
= 1.050 mol Cu
= 0.3500 mol P
= 1.401 mol O
/ 0.3500 mol
/ 0.3500 mol
/ 0.3500 mol
=3 Cu
= 1 P
= 4 O
Cu3PO4
66.75% Cu
10.84 % P
22.41 % O
66.75g Cu
10.84g P
22.41g O
€
1 mol Cu
63.55 g Cu
⎛
⎝ ⎜
⎞
⎠ ⎟
€
1 mol P
30.97 g P
⎛
⎝ ⎜
⎞
⎠ ⎟
€
1 mol O
16 g O
⎛
⎝ ⎜
⎞
⎠ ⎟
copper (I) phosphate
Step 1) % g Step 2) g mol Step 3) mol mol
Cu3PO4
X
X
X
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Empirical FormulaQuantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen.
Find the empirical formula of this compound.
= 1.408 mol Na
= 0.708 mol S
= 2.812 mol O
/ 0.708 mol
/ 0.708 mol
/ 0.708 mol
= 2 Na
= 1 S
= 4 O
Na2SO4
32.38% Na
22.65% S
44.99% O
32.38 g Na
22.65 g S
44.99 g O
Na g 23
Na mol 1
S g 32
S mol 1
O g 16
O mol 1
sodium sulfate
Step 1) % g Step 2) g mol Step 3) mol mol
Na2SO4
X
X
X