A student prepared 200mL of 0.01M solution of calcium phosphate, Can(PO4)2. What is the concentration of calcium in parts per million? Relative atomic masses Can =40, P =31, O =16
Bolaji, the molecular mass of the given compound is NOT 512. The compound calcium phosphate should be written as Ca3(PO4)2, and has a mass of 310 u.
You do not multiply the masses of calcium and phosphate, this is not an algebraic equation.
You add the masses of calcium and phosphate. (3*40 + 2*95)
As for the question, I am not sure of the exact methodology and it would be best to wait for a Jiskha chemistry tutor.
A student prepared 200mL of 0.01M solution of calcium phosphate, Can(PO4)2. What is the concentration of calcium in parts per million? Relative atomic masses Can =40, P =31, O =16.
compound: Ca3(PO4)2, molecular mass= (40*3+2*31+8*16)=310
so the total mass of calcium in the solution is .2*.01*120=.24 grams.
the total mass of the solution is 200 ml*1g/ml (it is a dilute solution), so you have
.24gramsCalcium/200 grams. So you want parts per million?
.24e6 milliongramsCa/200grams=240000/200=1200
check my math.
by the way calcium is Ca
Ca(PO^4)2
Ca=40, P=31, O=16
In BODMAS we deal with the ones in he bracket first
Ca (P O^4)2
40 (31 + 16*4)2
the reason why i multiplied 16 by 4 is because it is raised to the ^4
40 (64)2
In mathematics when you have a(b) it means a*b
40 (64*2)
40 (128)
the relative atomic mass is 512
but we are looking for for calcium which is 40 per million
note: when they say percent is over 100 because a cent is 100
40/512 * 1000000/1
=78125
To calculate the concentration of calcium in parts per million (ppm), we need to determine the number of moles of calcium in the given 200 mL solution and then convert it to ppm.
First, let's calculate the number of moles of calcium phosphate (Ca3(PO4)2) in the solution:
1. Calculate the molar mass of Ca3(PO4)2:
- Calcium (Ca) has a relative atomic mass of 40 g/mol.
- Phosphorus (P) has a relative atomic mass of 31 g/mol.
- Oxygen (O) has a relative atomic mass of 16 g/mol.
Molar mass of Ca3(PO4)2 = (3 * Ca) + (2 * P) + (8 * O)
= (3 * 40) + (2 * 31) + (8 * 16)
= 120 + 62 + 128
= 310 g/mol
2. Calculate the number of moles of Ca3(PO4)2 using the given volume and molarity:
Moles = Volume (in L) * Molarity
= 200 mL / 1000 mL/L * 0.01 mol/L
= 0.002 mol
Next, we need to determine the molar ratio of calcium to Ca3(PO4)2. From the formula Ca3(PO4)2, we can see that there are three calcium ions (Ca) for every one unit of Ca3(PO4)2.
3. Calculate the number of moles of calcium:
Moles of calcium = Moles of Ca3(PO4)2 * (3 moles of Ca / 1 mole of Ca3(PO4)2)
= 0.002 mol * 3
= 0.006 mol
Now, we will calculate the concentration of calcium in ppm:
4. Concentration of calcium in ppm = (Moles of calcium / Total volume in L) * 1000000
= (0.006 mol / 0.2 L) * 1000000
= 30,000 ppm
Therefore, the concentration of calcium in the 0.01M calcium phosphate solution is 30,000 parts per million (ppm).