Biology, asked by karishmakasarlawar51, 3 days ago

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



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Answered by Anonymous
4

Answer:

. 14 February it is truth there was something special on this day

.

.

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

.

Answered by Anonymous
4

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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