Copper is commonly mined as an ore with a variable percent composition of copper (II) sulfide. This ore is also sometimes referr
1 answer:
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Thus, given the 1.00-kg of 12.5% ore, we can compute the theoretical yield of copper (II) oxide via stoichiometry:
Whereas the third factor accounts for the percent purity of the covellite. Then, given the percent yield, we can compute the actual yield by:
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Answer: The molecular of the compound is,
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of
Mass of
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in of carbon dioxide,
of carbon will be contained.
For calculating the mass of hydrogen:
In 18 g of water, 2 g of hydrogen is contained.
So, in of water,
of hydrogen will be contained.
For calculating the mass of oxygen:
Mass of oxygen in the compound =
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Moles of Oxygen =
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is moles.
For Carbon =
For Hydrogen =
For Oxygen =
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 2 : 3 : 1
Hence, the empirical formula for the given compound is
The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :
Molecular formula =
Therefore, the molecular of the compound is,