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Question

Question: What mass is associated with a \( 2 \) mole quantity of potassium dichromate?...

What mass is associated with a 22 mole quantity of potassium dichromate?

Explanation

Solution

Potassium dichromate is an inorganic compound with the molecular formula of K2Cr2O7{K_2}C{r_2}{O_7} . The number of moles of any compound is the ratio of the mass and molar mass of that compound. By substituting the moles and molar mass given the mass of potassium dichromate.
n=wMn = \dfrac{w}{M}
nn is number of moles
ww is mass of potassium dichromate
MM is the molar mass of potassium dichromate.

Complete answer:
Inorganic compounds are the compounds that were formed from the atoms other than carbon and hydrogen. The molecular formula of potassium dichromate is K2Cr2O7{K_2}C{r_2}{O_7} which consists of two potassium atoms, two chromium atoms, and seven oxygen atoms. From all these numbers of atoms and their molar masses based on the periodic table.
Molar mass of potassium atoms is 40gmol140gmo{l^{ - 1}} , two potassium atoms has 2×40=80gmol12 \times 40 = 80gmo{l^{ - 1}}
Molar mass of chromium atoms is 52gmol152gmo{l^{ - 1}} , two chromium atoms has 2×52=104gmol12 \times 52 = 104gmo{l^{ - 1}}
Molar mass of oxygen atom is 16gmol116gmo{l^{ - 1}} , seven oxygen atoms has 7×16=112gmol17 \times 16 = 112gmo{l^{ - 1}}
Thus, the molar mass of potassium dichromate is 80+104+112=296gmol180 + 104 + 112 = 296gmo{l^{ - 1}}
Given that potassium dichromate is associated with 22 moles
Substitute the moles, and molar mass in the above formula,
2mol=w296gmol12mol = \dfrac{w}{{296gmo{l^{ - 1}}}}
Thus, mass of potassium dichromate is 296×2=592g296 \times 2 = 592g
592g592g of mass is associated with a 22 mole quantity of potassium dichromate.

Note:
Some atoms in the periodic table do not have the molar mass that is double to the atomic number. Thus, based on the periodic table only the molar mass of all the atoms in potassium dichromate should be calculated. Chromium and potassium molar masses are not equal to the double of the atomic number.