The electron configuration of a neutral elerpent is 1s 2s 2p 3s. If the electron configuration
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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>
Answer:
a) Percentage by mass of carbon: 18.3%
Percentage by mass of hydrogen: 0.77%
b) Percentage by mass of chlorine: 80.37%
c) Molecular formula:
Explanation:
Firstly, the mass of carbon must be determined by using a conversion factor:
The same process is used to calculate the amount of hydrogen:
The percentage by mass of carbon and hydrogen are calculated as follows:
%C= 18.3%
%H=0.77%
From the precipation data it is possible obtain the amount of chlorine present in the compound:
= 0.43g AgCl
Let's calculate the percentage by mass of chlorine:
%Cl== 80.37%
Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:
Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be
≈ 2
≈3
The empirical formula for the compound is:
The mass of this empirical formula is:
mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g
This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.