take out lab calculations & stampsheet . homework … read 8.8 #’s 54-58, 75, 76
DESCRIPTION
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
![Page 1: Take out lab calculations & stampsheet . Homework … Read 8.8 #’s 54-58, 75, 76](https://reader036.vdocuments.site/reader036/viewer/2022081503/56814fe7550346895dbdb338/html5/thumbnails/1.jpg)
Take out lab calculations & stampsheet.
Homework …1.Read 8.82.#’s 54-58, 75, 76
![Page 2: Take out lab calculations & stampsheet . Homework … Read 8.8 #’s 54-58, 75, 76](https://reader036.vdocuments.site/reader036/viewer/2022081503/56814fe7550346895dbdb338/html5/thumbnails/2.jpg)
Do Now:
If one stick of Juicy Fruit gum weighs 3.0g, what percent of the total mass of the gum is sugar?
![Page 3: Take out lab calculations & stampsheet . Homework … Read 8.8 #’s 54-58, 75, 76](https://reader036.vdocuments.site/reader036/viewer/2022081503/56814fe7550346895dbdb338/html5/thumbnails/3.jpg)
% 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
![Page 4: Take out lab calculations & stampsheet . Homework … Read 8.8 #’s 54-58, 75, 76](https://reader036.vdocuments.site/reader036/viewer/2022081503/56814fe7550346895dbdb338/html5/thumbnails/4.jpg)
Hydrate Lab
HydrateAnhydrous
CuSO4 ∙ ?H2O CuSO4
![Page 5: Take out lab calculations & stampsheet . Homework … Read 8.8 #’s 54-58, 75, 76](https://reader036.vdocuments.site/reader036/viewer/2022081503/56814fe7550346895dbdb338/html5/thumbnails/5.jpg)
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
![Page 6: Take out lab calculations & stampsheet . Homework … Read 8.8 #’s 54-58, 75, 76](https://reader036.vdocuments.site/reader036/viewer/2022081503/56814fe7550346895dbdb338/html5/thumbnails/6.jpg)
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
![Page 7: Take out lab calculations & stampsheet . Homework … Read 8.8 #’s 54-58, 75, 76](https://reader036.vdocuments.site/reader036/viewer/2022081503/56814fe7550346895dbdb338/html5/thumbnails/7.jpg)
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
![Page 8: Take out lab calculations & stampsheet . Homework … Read 8.8 #’s 54-58, 75, 76](https://reader036.vdocuments.site/reader036/viewer/2022081503/56814fe7550346895dbdb338/html5/thumbnails/8.jpg)
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
![Page 9: Take out lab calculations & stampsheet . Homework … Read 8.8 #’s 54-58, 75, 76](https://reader036.vdocuments.site/reader036/viewer/2022081503/56814fe7550346895dbdb338/html5/thumbnails/9.jpg)
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
![Page 10: Take out lab calculations & stampsheet . Homework … Read 8.8 #’s 54-58, 75, 76](https://reader036.vdocuments.site/reader036/viewer/2022081503/56814fe7550346895dbdb338/html5/thumbnails/10.jpg)
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
![Page 11: Take out lab calculations & stampsheet . Homework … Read 8.8 #’s 54-58, 75, 76](https://reader036.vdocuments.site/reader036/viewer/2022081503/56814fe7550346895dbdb338/html5/thumbnails/11.jpg)
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