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