The answer for the following problem is mentioned below.
- <u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.</em></u>
Explanation:
Given:
mass of calcium phosphate (
) = 125.3 grams
We know;
molar mass of calcium phosphate (
) = (40×3) + 3 (31 +(4×16))
molar mass of calcium phosphate (
) = 120 + 3(95)
molar mass of calcium phosphate (
) = 120 +285 = 405 grams
<em>We also know;</em>
No of molecules at STP conditions(
) = 6.023 × 10^23 molecules
To solve:
no of molecules present in the sample(N)
We know;
N÷
=
N =(405×6.023 × 10^23) ÷ 125.3
N = 19.3 × 10^23 molecules
<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules</em></u>
Knowing the ratio between atoms we can write an empirical formula:
<span>C4H6O </span>
<span>we compute the molar mass of this single formula: </span>
<span>4x12 + 6 x 1 + 16 x1 = 70 g / mol </span>
<span>Now, as we know the actual molar mas being 280 g/mol, we divide this number by 70 and we get the ratio between empirical formula and molecular actual formula: </span>
<span>280 / 70 = 4 </span>
<span>This means that actual molecular formula is: </span>
<span>(C4H6O)4 or </span>
<span>C16H24O4 </span>