How many total number of all the elements present in mole of (NH4)2 Cr2O7? Thanks a lot.

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How many total number of all the elements present in mole of (NH4)2 Cr2O7? Thanks a lot.

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Answer 1 · 2016-11-19T19:05:57

We use $Avogadro's number, N_{A} = 6.022 \times 10^{23} \cdot m o l^{- 1}$ $"Avogadro's number, "N_A=6.022xx10^23*mol^-1$ , to quantify the numbers of individual atoms in a mole, i.e. a $252.07 g$ $"252.07 g"$ mass of ${(N H_{4})}_{2} C r_{2} O_{7}$ $(NH_4)_2Cr_2O_7$ .......

Explanation:

In such a mass, there are $2 \times N_{A}$ $2xxN_A$ $nitrogen atoms$ $"nitrogen atoms"$ , and

$8 \times N_{A}$ $8xxN_A$ $hydrogen atoms$ $"hydrogen atoms"$ , and

$2 \times N_{A}$ $2xxN_A$ $chromium atoms$ $"chromium atoms"$ , and

$7 \times N_{A}$ $7xxN_A$ $oxygen atoms$ $"oxygen atoms"$ .

Here we use $N_{A}$ $N_A$ as a counting number, i.e. as we would use the number $10$ $10$ , or a $million$ $"million"$ , or a $dozen$ $"dozen"$ .

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How many total number of all the elements present in mole of (NH4)2 Cr2O7? Thanks a lot.

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