Commercial brass, an alloy of Zn and Cu, reacts with hydrochloric acid as follows.
Zn(s) + 2 HCl(aq) ZnCl2(aq) + H2(g)
(Cu does not react with HCl.) When 0.5093 g of a certain brass alloy is reacted with excess HCl, 0.0953 g ZnCl2 is eventually isolated.
(a) What is the composition of the brass by mass?
%Zn
%Cu
I have done this problem and got 32.42% Zn, 9.032% Zn, and 4.57% Zn and they were all wrong. I also got 35.15%Cu and that was wrong also.
8 answers
If you will show your work I will find the error.
I found the mass of ZnCl2 to be 136.29g
I started out:
.0953/136.3= 6.9e-4
then I did:
6.9e-4 x 65.39= .046 and multiplied that by a hundred
I started out:
.0953/136.3= 6.9e-4
then I did:
6.9e-4 x 65.39= .046 and multiplied that by a hundred
Two things wrong here.
1. You are rounding off incorrectly. First, you are allowed three significant figures and 0.0953/136.3 = 6.99E-4. Actually your are throwing away part of the answer.
2. Then 6.99E-4 x 65.39 = grams Zn and for that I have 0.0457g.
3. At this point you left out the last step. % Zn = (grams Zn/g sample) = (0.0457/0.5093)*100 = 8.977% which I would round to three s.f. as 8.98%
Then % Cu = 100% - %Zn
1. You are rounding off incorrectly. First, you are allowed three significant figures and 0.0953/136.3 = 6.99E-4. Actually your are throwing away part of the answer.
2. Then 6.99E-4 x 65.39 = grams Zn and for that I have 0.0457g.
3. At this point you left out the last step. % Zn = (grams Zn/g sample) = (0.0457/0.5093)*100 = 8.977% which I would round to three s.f. as 8.98%
Then % Cu = 100% - %Zn
i understand where my mistake was. when i went to find the %Cu i did:
100%-8.98%= 91.02 (rounding to three sig figs would give me 91.0)
both those answers were wrong on the on the online grader
100%-8.98%= 91.02 (rounding to three sig figs would give me 91.0)
both those answers were wrong on the on the online grader
I'm not privy to your on-line database but the USUAL thing wrong when they don't agree with your answer is the number of significant figures.
First, I assume you have a set of atomic masses and/or a periodic table that the database uses to calculate these things. Check those to make sure ZnCl2 and Zn are the same as you are using. I have a web site I use for these calculations and it gives me the same numbers you have used.
Second, I note that the grams for sample is to 4 places but the ZnCl2 is to 3 so that gives just 3 places to use for the answer. Check the numbers in your problem to make SURE you copied it correctly in your post.
If all of these check out ok make a note to talk to your prof tomorrow. From what you have posted, however, these are the right numbers assuming the database used the same values for molar masses.
First, I assume you have a set of atomic masses and/or a periodic table that the database uses to calculate these things. Check those to make sure ZnCl2 and Zn are the same as you are using. I have a web site I use for these calculations and it gives me the same numbers you have used.
Second, I note that the grams for sample is to 4 places but the ZnCl2 is to 3 so that gives just 3 places to use for the answer. Check the numbers in your problem to make SURE you copied it correctly in your post.
If all of these check out ok make a note to talk to your prof tomorrow. From what you have posted, however, these are the right numbers assuming the database used the same values for molar masses.
Okay, it needed four sig figs.
the next part of the question is:
How could this result be checked without changing the above procedure?
What does this mean?
the next part of the question is:
How could this result be checked without changing the above procedure?
What does this mean?
a.
If that is 0.0953 and not 0.09530g for ZnCl2, then you are allowed only 3 places.
b.
which result? Result for Zn or result for Cu. I can't think of another way without changing the procedure for Zn but for Cu you can do it another way this way.
g Zn = 0.0457
g sample = 0.5093
g Cu = 0.5093-0.0457 = ?
% Cu = (g Cu/g sample)*100 = ?
Of course this assumes everyone answered the Cu part by %Cu = 100=%Zn. Some students will always do it as I've shown the alternative in which case the shorter route would the the "other" way. :-)
Let me point out something else.
If you do it my alternative route, then four s.f. is the way to go; i.e.,
0.5093-0.0457 = 0.4636 (4 places are allowed doing it this way for Cu), then
(0.4636/0.5093)*100 = 91.0269 which I would round to 91.03% TO FOUR S.F. and my statement still is right---when the databases don't agree USUALLY it is a matter of s.f. :-).
If that is 0.0953 and not 0.09530g for ZnCl2, then you are allowed only 3 places.
b.
which result? Result for Zn or result for Cu. I can't think of another way without changing the procedure for Zn but for Cu you can do it another way this way.
g Zn = 0.0457
g sample = 0.5093
g Cu = 0.5093-0.0457 = ?
% Cu = (g Cu/g sample)*100 = ?
Of course this assumes everyone answered the Cu part by %Cu = 100=%Zn. Some students will always do it as I've shown the alternative in which case the shorter route would the the "other" way. :-)
Let me point out something else.
If you do it my alternative route, then four s.f. is the way to go; i.e.,
0.5093-0.0457 = 0.4636 (4 places are allowed doing it this way for Cu), then
(0.4636/0.5093)*100 = 91.0269 which I would round to 91.03% TO FOUR S.F. and my statement still is right---when the databases don't agree USUALLY it is a matter of s.f. :-).
Thank you so much!
4 answers