# How many atoms of Ca are present?

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A blood sample is found to contain 8.6 mg/dl of Ca. How many atoms of Ca are present in 8.6 mg? The atomic weight of Ca is 40.08 amu.

Cans someone help explain what I need to do to solve this problem? Thanks!

• One mole of anything, no matter what the element is, is Avogadro’s number, which is 6.02 * 10^23 atoms.

One mole of Calcium is equal to it’s atomic weight in grams :

1 mole Ca = 40.08 g = 6.02 * 10^23 atoms

There is a 1000mg in a gram. 1000mg * 40.08g =

=40,080mg of Ca per mole

If you have 8.6mg of Ca that is 8.6mg / 40,080mg = .00021457 of a mole of calcium

so just multiply that by Avogadro’s number to give you number of atoms:

.00021457 * 6.02 * 10^23 atoms = 1.292 * 10^20 atoms of Ca

• The question format here requires you to realize that MOLE calculations are required. I have chosen to answer this question using 40.08 amu = 1 mol Ca. Therefore, you must convert mg to grams , then calculate moles Ca, then solve for number of atoms using Avogadro’s number ( 6.023×10^23 atoms/mol)

1st convert 8.6 mg to grams by dividing by 1000 mg.

Next, convert that answer to mols by dividing the 40.08 amu value.

Next, take that answer and determine the number of atoms using Avogadro’s number by multiplying by (6.023 x 10^23 atoms/mol)

Solution:

8.6 mg Ca x (1 g/1000 mg)= 0.086 g Ca

0.086 g Ca x (1 mol Ca/ 40.08 g Ca)= 0.002145 mol Ca

0.002145 mol Ca x ( 6.023 x 10^23 atoms Ca/ mol Ca= 0.0129 x 10 ^23 = 1.29 x 10^21 atoms ofCa

Substituting this answer back into the original format gives you

1.29 x 10 ^ 21 atoms / dl of blood. (Don’t forget to round off if you are required to use significant figures in your answers!)

• there are (8.6 mg x 1g/1,000 mg x 1 mol/40.08g) mol Ca. Make sure you understand why this is.

each mol Ca contains Avogadro’s number of atoms of Ca.

the rest is just arithmetic.

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