A 110.0g sample of a mixture of CaCl2 and NaCl is treated with Na2CO3 to precipitate the Calcium as Calcium Carbonate. This CaCO3 is heated to convert all the Calcium into CaO and the final mass of CaO is 11.62 grams. The % by mass of the CaCl2 in the original mixture has to be found out..

i've tried out finding the no of moles of CaCO3 which which i got as 0.2075 and assuming that the CaCO3 and the no of moles of CaCl2 should be equal i tried finding out the % which came out to be 20.75% in this method and if i separately found the masses of Ca, Cl2 and summed them up to find the percentage over 110 the % turned out to be 20.09% need some suggestions to get the % anywhere near 15.2% in this process taking all of the compounds to be anhydrous

New contributor
kekule is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
  • 3
    $\begingroup$ Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers with units. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger permanent value. You may find useful formatting mathematical/chemical expressions/formulas. $\endgroup$
    – Poutnik
    May 25 at 18:31
  • $\begingroup$ Assuming no waste during the reaction process, amount of $\ce{CaO}$ isolated in $\pu{mol}$ is same as amount of $\ce{CaCl2}$ in $\pu{mol}$ in original mixture. You also have to assume $\ce{CaCl2}$ in anhydrous form during the calculation. $\endgroup$ May 25 at 20:19

1 Answer 1


There is no way to produce $11.62 \pu{g}$ of $\ce{CaO}$ from a $\ce{CaCl2 + NaCl}$ sample weighing $110.0 \pu{g}$ which is $15.2\%$ $\ce{CaCl2}$ by weight. Your calculations are correct.*

Assuming $100\%$ yield, and, as mentioned, no hydration whatsoever, you need at least $20.96\%$ $\ce{CaCl2}$ in a $110.0 \pu{g}$ $\ce{CaCl2 + NaCl}$ sample to produce $11.62 \pu{g}$ of $\ce{CaO}$.

*However, if the final product is hydrated, this is possible. Let me know if you need help with calculating percentage hydration.

Underlying Principle

This is an application of the well-known law of conservation of matter, also known as the principle of atom conservation (POAC). You may refer to R. C. Mukherjee (2004), Modern Approach to Chemical Calculations: An Introduction to the Mole Concept, 7$^\text{th}$ edition for a thorough reading with lots of practice problems.


Your overall (contracted) reaction is:

$$ \ce{CaCl2 -> -> CaO} $$

We simply apply POAC:

$$ \text{moles of }\ce{Ca}\text{ in }\ce{CaO} = \text{moles of }\ce{Ca}\text{ in }\ce{CaCl2} \tag{1} $$

$$ \text{moles of }\ce{Ca}\text{ in }\ce{CaO} = \dfrac{m_\ce{CaO}}{M_\ce{CaO}}M_\ce{Ca} = \dfrac{11.62}{56.077}40.078 \pu{g} = 8.30\pu {g} \tag{2} $$

$$ \text{moles of }\ce{Ca}\text{ in }\ce{CaCl2} = \dfrac{m_\ce{CaCl2}}{M_\ce{CaCl2}}M_\ce{Ca} = \dfrac{m_\ce{CaCl2}}{110.98}40.078 = 0.36m_\ce{CaCl2} \tag{3} $$

where, in Equations (2) and (3), $M_i$ is the molar mass of $i$ and $m_i$ is the mass of $i$ in sample of.

Equating Equations (2) and (3) in accordance with Equation (1):

$$ 8.30\pu {g} = 0.36m_\ce{CaCl2} \implies m_\ce{CaCl2} = 23.06 \pu{g} $$

Thus, if the original $\ce{CaCl2 + NaCl}$ sample weighed $110.0 \pu{g}$, $\ce{CaCl2}$ was present in $\dfrac{23.06}{110.0}\times100 \%$ weight ratio, which is $20.96\%$

Note: All calculations have been rounded off to the second decimal place.


Not the answer you're looking for? Browse other questions tagged or ask your own question.