Atoms to Grams

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February 2nd 2010, 02:45 PM
Nismo

Atoms to Grams

Hi, sry again -_- but i have yet another question.. what am i doing wrong!?

im trying to convert Atoms to Grams. EXample

1.0×1023 http://session.masteringchemistry.co...3B+atoms%7D%7D

so im doing

(1.0×1023 http://session.masteringchemistry.co...3B+atoms%7D%7D )(1/6.022x10^23)(196.
96655)=3.3g (putting it to 2 sig figs.)

same thing for 2.80×1022 http://session.masteringchemistry.co...3B+atoms%7D%7D

im doing (2.80×1022 http://session.masteringchemistry.co...3B+atoms%7D%7D )(1/6.022x10^23)(4.002602)=1.9

______________________________

last but not least can you guys help me get an idea of how to do this problem i think i know how but want to verify.

How many helium atoms are there in a helium blimp containing 535 http://session.masteringchemistry.com/render?units=kg of helium?

so i would start off by puting 535 into grams (535000g)then divide by molar mass then multiply by avogadro = atoms (8.0*10^28)
February 2nd 2010, 02:55 PM
e^(i*pi)

Quote:

Originally Posted by Nismo

Hi, sry again -_- but i have yet another question.. what am i doing wrong!?

im trying to convert Atoms to Grams. EXample

1.0×1023 http://session.masteringchemistry.co...3B+atoms%7D%7D

so im doing

(1.0×1023 http://session.masteringchemistry.co...3B+atoms%7D%7D )(1/6.022x10^23)(196.
96655)=3.3g (putting it to 2 sig figs.)

same thing for 2.80×1022 http://session.masteringchemistry.co...3B+atoms%7D%7D

im doing (2.80×1022 http://session.masteringchemistry.co...3B+atoms%7D%7D )(1/6.022x10^23)(4.002602)=1.9

______________________________

last but not least can you guys help me get an idea of how to do this problem i think i know how but want to verify.

How many helium atoms are there in a helium blimp containing 535 http://session.masteringchemistry.com/render?units=kg of helium?

so i would start off by puting 535 into grams (535000g)then divide by molar mass then multiply by avogadro = atoms (8.0*10^28)

I'd use the mole as an intermediate step, that is divide by Avogadro's constant and then by the relative atomic mass.

$\text { moles of gold} = \frac{10^{23}}{6.02 \times 10^{23}} = \frac{1}{6.02}$

$m = n\,A_r = \frac{1}{6.02} \times 197 = 32.72 \text{ g }$

You appear to be an order of magnitude out by my answers, check your powers of 10?
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2. Yep
February 2nd 2010, 03:19 PM
Nismo

1) I see what you did for the 1st problem, but i cant seem to use that method when doing

7.4×1021 http://session.masteringchemistry.co...3B+atoms%7D%7D or 2.80×1022 http://session.masteringchemistry.co...3B+atoms%7D%7D

2) Also is the answer going to be 8 atoms or the 8.049x10^24?? (because i would have to round this to 4 sig figs since the 6.022 is 4 sig figs and since im dividing with if that makes sense?