Concept explainers
(a)
Interpretation:
The empirical formula of the compound containing
Concept Introduction:
Empirical Formula:
The empirical formula of a compound is the simplest whole number ratio of each type of atom in a compound. It can be the same as the compound’s molecular formula but not always. An empirical formula can be calculated from information about the mass of each element in a compound or from the percentage composition.
The steps for determining the empirical formula of a compound as follows:
- Obtain the mass of each element present in grams.
- Determine the number of moles of each atom present.
- Divide the number of moles of each element by the smallest number of moles.
- Convert the numbers to whole numbers. The set of whole numbers are the subscripts in the empirical formula.
(a)
Answer to Problem 37PE
The empirical formula of the compound is
Explanation of Solution
Given,
The grams of zinc is
The grams of carbon is
The grams of oxygen is
The
The atomic mass of zinc is
The atomic mass of oxygen is
The grams of each element has to be converted to moles as,
The moles of carbon
The moles of Zinc
The moles of oxygen
The empirical formula can be calculated as,
The number of moles can be converted to moles to whole numbers by dividing by the small number.
The empirical formula of the compound is
(b)
Interpretation:
The empirical formula of the compound containing
Concept Introduction:
Refer to part (a).
(b)
Answer to Problem 37PE
The empirical formula of the compound is
Explanation of Solution
Given,
The grams of the compound is
The grams of carbon is
The grams of hydrogen is
The grams of chlorine is
The atomic mass of carbon is
The atomic mass of hydrogen is
The atomic mass of chlorine is
The grams of each element has to be converted to moles as,
The moles of carbon
The moles of hydrogen
The moles of Chlorine
The empirical formula can be calculated as,
The number of moles can be converted to moles to whole numbers by dividing by the small number.
The empirical formula of the compound is
(c)
Interpretation:
The empirical formula of the compound containing
Concept Introduction:
Refer to part (a).
(c)
Answer to Problem 37PE
The empirical formula of the compound is
Explanation of Solution
Given,
The grams of the compound is
The grams of vanadium is
The grams of oxygen is
The atomic mass of vanadium is
The atomic mass of oxygen is
The grams of each element has to be converted to moles as,
The moles of vanadium
The moles of oxygen
The empirical formula can be calculated as,
The number of moles can be converted to moles to whole numbers by dividing by the small number.
Since the value is fractional multiply the value by two. The empirical formula of the compound is
(d)
Interpretation:
The empirical formula of the compound containing
Concept Introduction:
Refer to part (a).
(d)
Answer to Problem 37PE
The empirical formula of the compound is
Explanation of Solution
Given,
The grams of nickel is
The grams of oxygen is
The grams of phosphorus is
The atomic mass of nickel is
The atomic mass of oxygen is
The atomic mass of phosphorus is
The grams of each element has to be converted to moles as,
The moles of nickel
The moles of oxygen
The moles of phosphorus
The empirical formula can be calculated as,
The number of moles can be converted to moles to whole numbers by dividing by the small number.
Since the value is fractional multiply the value by two. The empirical formula of the compound is
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Chapter 7 Solutions
FOUND.OF COLLEGE CHEMISTRY
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