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Lecture 3

CHEM 131 Lecture Notes - Lecture 3: Molar Mass, Empirical Formula, Chemical Formula


Department
Chemistry
Course Code
CHEM 131
Professor
Ladon
Lecture
3

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Chapter 3 - Stoichiometry
Basic stoichiometry
C + O2 => CO2
Reactants Product
12 amu 32amu 44amu
1mol C 2mol O 1mol CO2
12g 32g 44g
!"#$%&
!#$%& x !#'
(.*"+!*,-#$%& π‘₯# (.*"+!*,-#/#$01%
!#$01% = !"#'
!#%12
Radioactive Decay Reaction
𝐿𝑖
5"
"67 -> π‘‡β„Ž
5*
"6: + π‘Ž
"
: alpha decay (add the two)
π‘‡β„Ž
5*
"6: -> π‘ƒπ‘Ž
5!
"6: + 𝐡
>!
* beta decay (minus the two)
Molar mass
Molar mass – number of grams of a substance and one mole of that substance
Mg(NO3)2 1 Mg x 24.3g/mol = 24.31g/mol
2 N x 14.01g/mol = 28.02g/mol
6 O x 16.00g/mol = 96.00g/mol
= 148.33g/mol
Meaning 1 mol Mg(NO3)2 = 148.33g
β€’!How many mole of Mg(NO3)2 are found in 15.73g Mg(NO3)2 ?
15.73g Mg(NO3)2 x !#%12#?' @A-"
!:7.66#' @A-" = 0.1060 mol Mg(NO3)2
β€’!How many grams of Mg(NO3)2 should be measure to have 0.517 mol Mg(NO3)2 ?
0.517 mol Mg(NO3)2 x !:7.66'#?' @A-"
!#%12#?' @A-" = 76.7 g Mg(NO3)2
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