Map ng learning Determine the molecular mass and its uncertainty for CH3CH2COOH.
ID: 978895 • Letter: M
Question
Map ng learning Determine the molecular mass and its uncertainty for CH3CH2COOH. Note: Significant figures are graded for this problem. To avoid errors, do not round your answers until the very end of your calculations and use the atomic masses given in the table to the right. Element Atomic Mass Number Number 1.00794 t 7 14.07854 g/ mol 12.0107 t 8 g/mol 15.9994 t 3 There is additional feedback availablel View this feedback by clicking on the bottom divider bar again to hide the additional feedback. Incorrect. Explana "AO Previous Try Again Next ExtExplanation / Answer
1)CH3CH2COOH or C3H6O2
Molar mass=(12.0107+/-8)*3+(1.00794+/-7)*6+(15.9994+/-3)*2
Use rule for propagation of errors for multiplication with a constant
R=X*c
R=X*lCl
For R+R=c*(X+/-X)
Molar mass=(36.0321 +/- 24)+ (6.04764+/-42)+(31.9988+/-6)
Use rule for propagation of errors for addition/substraction
R=X+Y
R=[(X)^2+(Y)^2]^1/2
For R=(X+/-X)+( Y+/-Y)
Molar mass=74.07854 +/-R
Where R=[(24)^2+(42)^2+(6)^2]^1/2=[576+1764+36]^1/2=48.7442
Molar mass=74.07854+/-48.7442(answer)
2)
A)(5.1)±(0.3)-3.7(±0.2)=
Use rule for propagation of errors for addition/substraction
R=X+Y
R=[(X)^2+(Y)^2]^1/2
For R=(X+/-X)+( Y+/-Y)
(5.1)±(0.3)-3.7(±0.2)=(5.1-3.7)±[(0.3)^2+(0.2)^2]^1/2=1.4±[0.36]
=1.4±0.36±relative error
%Rel error=absolute error/measured quantity*100
Or for R±R rel error=R/R*100
So % rel error=0.36/1.4*100=25.71
Answer=1.4±0.36±25.71%
B)(8.24)±(0.04)*0.012±(0.001)
Use Use rule for propagation of errors for multiplication/division
R=X*Y
R=lRl*[(X/X)^2+(Y/Y)^2]^1/2
8.24)±(0.04)*0.012±(0.001)=8.24*0.012±[(0.04/8.24^2+(0.001/0.012)^2]^1/2=0.09888±[0.0000235+0.006889]^1/2=0.09888±0.0±0.08314/0.09888*100
=0.09888±0.08314 ±84.08%
C) same as A and B
D)9.2(±0.7)*([5.4(±0.3)*10^-3]+([5.6(±0.1)*10^-3]
Use rule for propagation of errors for multiplication/division
And
Use rule for propagation of errors for multiplication with a constant
9.2(±0.7)*([5.4(±0.3)*10^-3]+([5.6(±0.1)*10^-3]
=9.2(±0.7)*([0.0054±0.0003]+[0.0056±0.0001]= 9.2(±0.7)*([ [0.011±(0.0003)^2+(0.0001)^2]^1/2
=9.2(±0.7)*[0.011±(0.00000009+0.00000001)^1/2=9.2(±0.7)*(0.011±0.000316)
=(9.2*0.0011)±[(0.7/9.2)^2+(0.000316/0.011)^2]^1/2=0.01012±[0.00577+0.000823]1/2
=0.01012±0.0812±0.0812/0.01012*100
=0.01012±0.0812±802.371%(answer)