take out lab calculations & stampsheet . homework … read 8.8 #’s 54-58, 75, 76
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Take out lab calculations & stampsheet . Homework … Read 8.8 #’s 54-58, 75, 76. Do Now : If one stick of Juicy Fruit gum weighs 3.0g, what percent of the total mass of the gum is sugar?. 24.305. 35.453. Mg. Cl. 12. 17. magnesium. chlorine. Percentage Composition. - PowerPoint PPT PresentationTRANSCRIPT
Take out lab calculations & stampsheet.
Homework …1.Read 8.82.#’s 54-58, 75, 76
Do Now:
If one stick of Juicy Fruit gum weighs 3.0g, what percent of the total mass of the gum is sugar?
% Cl = x 10070.90 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
25.53% Mg
74.47% Cl
(by mass...not atoms)
It is not 33% Mg and 66% Cl
% Mg = x 10024.31 g95.21 g
Hydrate Lab
HydrateAnhydrous
CuSO4 ∙ ?H2O CuSO4
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
Empirical and Molecular Empirical and Molecular FormulasFormulas
A pure compound always consists of the same elements combined in the same proportions by weight.Therefore, we can express
molecular composition as PERCENT PERCENT BY WEIGHTBY WEIGHT.Ethanol, C2H6O
52.13% C 13.15% H 34.72% O
Different Types of FormulasMolecular 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
The Empirical Formula The Empirical Formula MarchMarch
Percent to massPercent to mass Mass to moleMass to mole Divide by smallestDivide by smallest Return to wholeReturn to 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
C g 12.01
C mol 1
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) return to whole
x2
x2
x2
X
X
X
Empirical FormulaQuantitative analysis shows that a compound contains 66.75% copper, 10.84% phosphorus and 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
Step 1) % g Step 2) g mol Step 3) mol mol
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
Step 1) % g Step 2) g mol Step 3) mol mol
X
X
X