Answered

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2. What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% 0, and a molecular mass of 1942 I Rre learned​

Sagot :

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Answer:

[tex]C_{8} H_{10} N_{4} O_{2}[/tex]

Explanation:

Mass of C in compound = 49.47 % of 194.2 amu  

                                    = 96.07074 amu

No. of moles of C in compound = 96,07074/12 = 8

Similarly,  Mass of H = 10 amu

No. of moles of H = 10/1 = 10

Mass of N = 56 g

No. of moles of N = 56/ 14 = 4

Mass of O = 32 g

No. of moles of O = 32/16 = 2

Hence, Molecular Formula - [tex]C_{8} H_{10} N_{4} O_{2}[/tex]