do now: if one stick of juicy fruit gum weighs 3.0g, what percent of the total mass of the gum is...
TRANSCRIPT
Do Now:
If one stick of Juicy Fruit gum weighs 3.0g, what percent of the total mass of the gum is sugar?
Hydrate Lab – for tomorrow
Pre-lab Title Purpose Materials Procedure Data table
Hydrate Anhydrous
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.
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
% 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
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
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
Calculating Empirical formula
Percent to mass (assume 100g) Mass to mole (molar mass) Divide by the smallest (ratio)
Multiply til’ whole*
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
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
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