11 grams

by Remi Leano | Fall 2019 ocf.berkeley.edu/~rleano

CHEM 001 MIDTERM 2 Blinky GuideCHAPTERS 7

,8

,9

,

10 ( WORKSHEETS 7 - 70,

QUIZZES 5 - 7)

GRAMS TO MOLES TO ATOMSPRACTICE PROBLEMS : W 6 ( SECTION VI ) P1

,W7P3

,W 8 PI

Ex.

?

18.02gramsmolar mass of H2O : 2 ( I

Iom},JIT ) -11 ( 16199048ms) =

I moi H2O

zzgrams H2o ×

I mot H2O×

6.022×10"

molecules H2O× z3mofetgmef-zo-1.la x 1024 atoms

18.02 gramsI mot H2O

PERCENT COMPOSITIONPRACTICE PROBLEMS : Q5P2

,W 7 P2

molar mass of Mgsoy : If 34in? g) t I ( Im:9% ) -14/7%981=139?m7g§o,

4 (' I'°m%)

= 0.5317 ⇒ 53.17%of the mass of Mg 504

(z"mo93mg7§q) isfromoxygen .

( 0.5317 ) ( 212.07g) = 112.8g of oxygen

by Remi Leano | Fall 2019 ocf.berkeley.edu/~rleano

MOLECULAR & EMPIRICAL FORMULAS

PRACTICE PROBLEMS : Q5P3,

Wb ( SECTION VI ) P2,

W7P4,

W7P5

$8000002.03

000063.57DRAFTAt.lk#

molar mass of C : 12.01g 11 mold

miiiiiisissoiti.is?oioo:iiiiii: ①I

(63.57g4/42.01glmolc) =

5.293mot C 11.985 = 2.666 x 3 = 7.998 → 8 C

(

31.76g01/46.00glmolO ) = 1.985 mot0/1.985 = 1 x 3 = 3 →

30(

4.67gHI / ( 1.01

glmolH ) =

4.624mot H 11.985 =

2.329×3=6.987

→ 7 H

empirical : Coo H7O3

molar mass of Cotta : 8( ' fifty ) -1 7 ( Iom! ,If ) -13/7%980) =

151.15g1 mot Coo # 703

302.3 glmol = 2 ⇒ molecular : Cho Hey 06151.15g Imd

by Remi Leano | Fall 2019 ocf.berkeley.edu/~rleano

O REACTIVITY SERIESpractice PROBLEMS : W8P3

YOU DO NOT NEED TO MEMORIZE

THE SERIES .

REMEMBER,

ANY NEW COMPOUNDS FORMED MUST BE ELECTRICALLY NEUTRAL .

EACH My IS 2 +

⇒ My ( NO 3)z

IS NEUTRAL

EACH N Oz IS 1-

oREACTION TYPES

practice problems : W8P2

S - - REACTANT TO PRODUCTpmtamcepropsu.ms

:Q6P1,w8P5.w9P1

4 2

molar mass of Sify : 14%99%1+417'm9f¥) = IT!%i¥,

molar mass of HF : I ( ' Iom?,¥ ) + I- ( ' I'Omg're ) = 3997¥

37.9g Site,

xt mot Sify

×4 mot HF

104.09g I moi sie ,

× 21%7It = 29.1g HE

by Remi Leano | Fall 2019 ocf.berkeley.edu/~rleano

LIMITING REACTANT TO PRODUCT & % YIELD

PRACTICE PROBLEMS : QGPZ,

W8P4,

W8( ALA ) ,

W9P2

THE DIFFERENCE WITH THE PREVIOUS PROBLEM IS THAT HERE WE ARE GIVEN QUANTITIES

OF TWO REACTANTS INSTEAD OF JUST ONE,

SO WE HAVE

TO FIND THE LIMITINGREACTANT .

7.5 4 5

multiply all by 2 to remove decimalSTEP I :

BALANCE 2 13 8 10

molarmass of Oz : 2 ( If.

moggy ) = 32.00g1 mot Oz

STEP 2 :

FIND LIMITING molar mass of Cy Hao : 4 ( Ying'T ) -110 (zoom! ,'t ) = I'm?Y&µoREACTANT

16.05g Cy Hyo xt mot 04h10

×8 mol coz

58.14g 2 mo , CyHyo= 1.104 mot CO2

32.00g Oz xt mot Oz

×8 mot coz

32.00g 13 moi Oz= 0.6154 mot CO2 tIf¥aI¥f

STEP 3 :molar mass of CO2 : 2116in?! ) -11/12.mg#)=I4m.ofzgfz

FIND GRAMS

OF PRODUCT 0.6154 mot CO2 ×

4140mg, a

= 27.08g CO2

STEP 4 :

APPLY Yo YIELD (27.08g CO2 ) ( 0.850 ) =

23.0g CO2

by Remi Leano | Fall 2019 ocf.berkeley.edu/~rleano

You do not need to know how to find the valence electrons for transition metals!

VALENCE ELECTRONS

PRACTICE PROBLEMS : Q7P5 , WIOPI

S

:15252106353104 SULFUR HAS 6 VALENCE ELECTRONS .

ELECTRON CONFIGURATIONPRACTICE PROBLEMS : Q7P1

,Q7P3

,W10P4 ,

WTOP 6

Keetoowah

P :

15252,5353ps

( add 3 electrons ) P'

:

15252106353106ORBITAL DIAGRAM

PRACTICE PROBLEMS : QFPI,

W10P3,

W10P4,

W10P5

P :

15252136353psP : ftttf-1 ItI I I

15 25 2ps35 3ps