Empirical formula of a compound if a 50.0 g sample of it contains 9.1g Na ,20.6 g Cr ,and 22.2 g O
1 answer:
Answer:
The empirical formula of a compound is Na2Cr2O7
Explanation:
Empirical formula is defined as the lowest ratio of elements present in any substance. In order to determine the empirical formula for a substance containing these elements we have to follow some steps.
1.In the first step, we have to divide the given mass of an element with its molecular mass to convert the mass to moles.
Similarly, the moles of Cr and O in that compound were calculated and found as 0.396 and 1.3875, respectively.
2. Then the next step is to determine the lowest moles and divide all the moles with that lowest mole. As here the lowest mole is 0.396. So on dividing the moles of Na, Cr and O with 0.396, we will obtain the ratio as 1:1:3.5.
3. As the ratio consists of a fractional number, we have to multiply it with the smallest possible integer to convert it to whole number. In this case, if we multiply it with 2 , we will get the ratio as 2:2:7.
Thus, the empirical formula for the compound can be written as Na2Cr2O7.
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Taking into account the reaction stoichiometry, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
CaO + H₂O → Ca(OH)₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CaO: 1 mole
- H₂O: 1 mole
- Ca(OH)₂: 1 mole
The following rule of three can be applied: If by stoichiometric reaction 1 mole of Ca(OH)₂ is produced by 1 mole of CaO, 2 moles of Ca(OH)₂ are produced by how many moles of CaO?
moles of CaO= 2 moles
Finally, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.
Learn more about the reaction stoichiometry:
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