What are the moles OR masses associated with: (i) #15.27*g*"lithium metal"#; (ii) #2.40*mol*"sulfur"#; (iii) #0.0141*mol# Ar; (iv) #88.0*mol*Mg#; (v) #2.29*g*"phosphorus"#; (vi) #11.9*mol*"chromium"#; (vii) #9.62*g;# (viii) #237.6 mol As#?

2 Answers
Jun 12, 2017

Answer:

#"Molar quantity"# #=# #"Mass"/"Molar mass"#

Explanation:

This is the basis, and we use this quotient to approach all of the given problems. I trust that you do have a Periodic Table beside you, because you need it to determine the atomic mass of each component atom. No-one could be expected to remember all the atomic masses, mind you, you will soon recall the common ones........

#15*g# #Li#, #"Moles"=(15*g)/(6.94*g*mol^-1)=2.16*mol#. And note that the quotient gave us a dimensionally consistent answer: we wanted an answer in moles, and thus #1/(1/"mol")=mol# as required.

#2.4*mol# #S#, #"Mass"=2.4*molxx32.06*g*mol^-1=76.9*g#. Was this dimensionally consistent?

#22*g# #Ar#, #"Moles"=(22*g)/(39.9*g*mol^-1)=0.551*mol#.

I could of course go on, but why don't you try to answer some yourself. We (I and others) can review your answers if you like.

Jun 12, 2017

Answer:

  1. #2.2# #"mol Li"#

  2. #77# #"g S"#

  3. #0.55# #"g Ar"#

  4. #2.14xx10^3# #"g Mg"#

  5. #0.074# #"mol P"#

  6. #619# #"g Cr"#

  7. #0.24# #"mol Ca"#

  8. #1.78xx10^4# #"g As"#

Explanation:

Each one of these problems involves the interconversion of grams and moles. You can reference a periodic table (or look them up online) for each element's molar mass (the atomic mass under the symbol, typically).

I'll work through one problem going from grams to moles, and one going from moles to grams, but the other six are practically the same process, so I'll just write the dimensional analysis.

1.

We're asked to find how many moles of lithium are in a sample with a mass of #15# #"g"#. The molar mass of lithium is, from a periodic table, #6.94 "g"/"mol"#. Using dimensional analysis, we have

#15cancel("g Li")((1"mol Li")/(6.94cancel("g Li"))) = 2.2# #"mol Li"#

rounded to #2# significant figures, the amount given in the question.

2.

This question is asking to convert from moles to grams, which is almost the exact same process. The molar mass of #"S"# is #32.07"g"/"mol"#, so

#2.4cancel("mol S")((32.07"g S")/(1cancel("mol S"))) = 77# #"g S"#

rounded to #2# significant figures.

The remaining problems are done similarly:

3.

#22cancel("g Ar")((1"mol Ar")/(39.95cancel("g Ar"))) = 0.55# #"g Ar"#

(#2# sig figs)

4.

#88.1cancel("mol Mg")((24.30"g Mg")/(1cancel("mol Mg"))) = 2.14xx10^3# #"g Mg"#

(#3# sig figs)

5.

#2.3cancel("g P")((1"mol P")/(30.97cancel("g P"))) = 0.074# #"mol P"#

(#2# sig figs)

6.

#11.9cancel("mol Cr")((52.00"g Cr")/(1cancel("mol Cr"))) = 619# #"g Cr"#

(#3# sig figs)

7.

#9.8cancel("g Ca")((1"mol Ca")/(40.08cancel("g Ca"))) = 0.24# #"mol Ca"#

(#2# sig figs)

8.

#238cancel("mol As")((74.92"g As")/(1cancel("mol As"))) = 1.78xx10^4# #"g As"#

(#3# sig figs)

Hope this helped:)