Question

Cuantos gramos de carbono hay en 1.25*1022 moleculas de éter dietílico

Answer 1

Mm (C2H5)2O = 74 g/mol

C: 4 X 12 = 48 g/mol

H: 10 x 1 = 10 g/mol

O: 1 x 16 = 16 g/mol

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      Mm = 74 g/mol

1. calcular moles de eter dietilico

     1 mol ---- 6.022 x 10²³ moléculs

         x    ---- 1.25 x 10²² moléculas

               x =  0.02075 mol de éter dietílico

2. calcular gramos de C

g C = 0.02075 mol C4H10O  x     48 g C

                                                   ````````````````````````

                                                     1 mol C4H10O

g C = 0.996 ≈ 1.00