Hello, Liz,
Please note my observations/questions:
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0.2 CaCO3 x 1mol CaCO3 / 100.09 g CaCO3 x = mole CaCO3 is correct
6.02x10^23 molecules CaCO3 / 1mol CaCO3 x3 = The conversion factor is correct, but why add the "3?". It should not be there.
5 atoms / 1 molecule CaCO2 = Does the question want to total number of atoms, or the total number of CaCO3 molecules? If molecules CaCO3, we don't need to use this conversion factor. It is correct, but onl;y if we wants total atoms, not CaCO3 molecules.
I got 6.017 x 10^21 atoms with sig fig, I think it should be round to the unit place? =You only have 1 sig fig in the 0.2 g, so the answer should only have 1 sig fig.
6 x 10^21 atoms? = I get 1.2 x 1021 molecules of CaCO3.
Bob
Liz B.
Oh, sorry, I understand your answer if up to finding the number of molecules of CaCO3. I still need to find the "total number of atoms" used. So I need to convert your answer in moleucles to atoms. So your answer 1.2 x 10^21 molecules of CaCO3 multiply 5 atoms / 1 molecule CaCO3 to get total number of atoms, and if my original answer is correct?01/26/21
Robert S.
01/27/21
Liz B.
Thank you so much. Is my answer suppose to round to 1 sig fig or 3?01/27/21
Liz B.
When I multiply and divide the conversion factors like the 6.02 x 10^23....Should I ignore the conversion factor when rounding sig fig? Let's say my original weight of the substance used is 0.2 g, my answer round to 1 sig fig? What if my original used amount is 0.21212g and my molar mass of CaCO3 round to 100.1g, my answer will be rounded to 4 sig fig? Does the molar mass conversion factor take into consideration? How about calculating the molar mass, when I have 1 mol of Ca at 40.08g, 1 mol of C at 12.011 g.... when I "add" all the mass together, should I round the molar mass to the lease precise value which is the hundredths?01/27/21
Robert S.
01/27/21
Liz B.
Hi Bob, Thank you so much for your help. The more I read, I'm getting confused again. For the same question, convert the 0.20 g of CaCO3 that I used to number of atoms. Some books or tutors told me this: 0.20 g CaCo3 x 1 mol CaCO3 / 100.09 g CaCO3 x 6.02 x 10^23 atoms / 1 mol CaCo3. Which is 1.2 x 10^25 atoms. My text book has this example: How many atoms are in 2.12 mol of C3H8? 2.12mol C3H8 x 6.02x10^23 molecules C3H8/1 mol C3H8 x 11atoms/1molecules C3H8 =1.40 x 10^25 atoms That's why when I convert 0,20g of CaCo3 to atoms, I convert it to moles , then to molecules, then atoms. (just like the previous post) I'm getting confused which is correct and when I can convert a compound from gram ---> moles ---> atoms by multiply 6,02x10^23 atoms/1mol And if converting a compound from grams ---> moles--> molecules-->atoms is correct? If so, how to determine when I need an extra step to convert to molecules then to atoms? Thank you for your help!01/27/21
Liz B.
Sorry, I think i have typo. My original should be: 0.2g CaCO3 x 1 mol CaCO3 / 100.09g CaCO3 x 6.02x10^23 molecules CaCO3 / 1 mol CaCO3 x 5 atoms / 1 molecules CaCO3 The question is just says find the "total number of atoms" used. Based on 0.2g CaCO3 is used. When I multiply and divide, I still got 0.06015 x 10^23 So 6.015 x 10^21 I can't get the same as yours.01/26/21