An analytical chemist analyzed a sample of a compound that is found to contain 31.25% Calcium, 18.75% Carbon and 50.0% Oxygen. What is the empirical formula of the compound? How about the molecular formula? The molar mass is 128 g/moli
21 July
and you
Answers
Answer:
. 14 February it is truth there was something special on this day
.
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Empirical formula is therefore Na
2
SH
20
O
14
empirical mass=2×23+32+20×1+14×16=322g
Given:
Molecular mass = 322 g
∴ molecular formula =empirical formula=Na
2
SH
20
O
14
Given that all the hydrogen in the compound is reset in combination with oxygen as water of crystallisation
If water of crystallization are nH
2
O then, 2n=20
or n=10
Crystallized water is 10H
2
O remaining
.
Step I : Calculation of simplest whole number ratios of the elements <br>
{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",34.6,12,(34.6)/(12)=2.88,(2.88)/(2.88)=1,3),("H",3.85,1,(3.85)/(1)=3.85,(3.85)/(2.88)=1.337or 4//3,4),("O",61.55,16,(61.55)/(16)=3.85,(3.85)/(2.88)=1.337or 4//3,4):}
<br> The simplest whole number ratios of the different elements are :
C : H : O : : 3 : 4: 4
<br> Step II. Writing the empirical formula of the compound. <br> The empirical formula of the compound
= C_(3)H_(4)O_(4)
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