BR Gen Chem Example 1.13

[Ihavesomanyquestions]

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BR Gen Chem Example 1.13


Which of the following 1.0-gram samples yields the MOST moles of sodium cation?

A. 1.0 g NaCl
B. 1.0 g NaBr
C. 1.0 g NaNO3
D. 1.0 g Na2CO3

Before looking at the answer explanation I calculated the values. What is wrong with my thinking. For example for answer choice A:

1.0 g NaCl (mol NaCl/58 g NaCl) (mol Na/mol NaCl)

Wouldn’t this calculation yield the number of moles?





Wouldn’t this calculation yield the number of moles?

 

[sillyjoe]

NaCl --> Na + Cl (1:1 ratio of Na reactant : Na product) you start with 1/58th of a mole [1/58 * 1 = 1/58th mole Na+]

NaBr --> Na + Br (1:1 ratio) you start with 1/103th of a mole [1/103 * 1 = 1/103th of a mole]

NaNO3 --> Na + NO3 (1:1 ratio) 1/85th of a mole Na

Na2CO3 --> 2 Na + CO3 (1:2 ratio) [1/106 * 2 = 1/53 mole of sodium]

This could also have been solved figuring out which has the greatest mass % of sodium since it is all 1 gram of salt you can find the largest mass % and equate that with the largest % of Na in 1 gram.

 

[dboyhaaan]

NaCl --> Na + Cl (1:1 ratio of Na reactant : Na product) you start with 1/58th of a mole [1/58 * 1 = 1/58th mole Na+]

NaBr --> Na + Br (1:1 ratio) you start with 1/103th of a mole [1/103 * 1 = 1/103th of a mole]

NaNO3 --> Na + NO3 (1:1 ratio) 1/85th of a mole Na

Na2CO3 --> 2 Na + CO3 (1:2 ratio) [1/106 * 2 = 1/53 mole of sodium]

This could also have been solved figuring out which has the greatest mass % of sodium since it is all 1 gram of salt you can find the largest mass % and equate that with the largest % of Na in 1 gram.

Sorry for bringing up an old thread I just assumed that D was the correct answer because its the only choice with a 2:1 or sodium : to rest of compound. Was I just lucky?

 

[BerkReviewTeach]

Sorry for bringing up an old thread I just assumed that D was the correct answer because its the only choice with a 2:1 or sodium : to rest of compound. Was I just lucky?

Technically you need to do it the above route if you were going for an exact answer in a general chemistry class. But for MCAT reasoning, your approach is time efficient and unless the anion associated with the two sodium species is abnormally massive, you will get the right answer your way.

 

[ProspectiveKidd]

Sorry for bringing up an old thread I just assumed that D was the correct answer because its the only choice with a 2:1 or sodium : to rest of compound. Was I just lucky?

Absolutely pissening. I used the same methodology.

 

[dboyhaaan]

Technically you need to do it the above route if you were going for an exact answer in a general chemistry class. But for MCAT reasoning, your approach is time efficient and unless the anion associated with the two sodium species is abnormally massive, you will get the right answer your way.

Ok, thanks I wasn't sure because it wasn't mentioned in the solution.

Absolutely pissening. I used the same methodology.

Not in the AM bro.


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