How many grams would 3.36 × 1023 molecules of copper (II) sulfate (CuSO4) weigh?

Question

thomaster · Answer

Divide 3.36 × 10^23 by avogadro's number (6.022 x 10^23)
Then you have the amount of moles copper(II)sulfate
You multiply that with the molecular weight of CuSO4 (159.6095 g/mol)

anonymous · Answer

mole = number of molecules/Avogadro number
mole = 3.36 x 10^23 / 6.02 x 10^23 =0.56
to find mass you multiply moles with relative molecular mass of CuSO4
mass = 0.56 x 159.5 = 89.32 grams

thomaster · Answer

$$\frac{ 3.36 * 10^{23} }{ 6.022 * 10^{23} }=0.5579 mol$$0.5579 * 159.6095 = 89.0461 gram

anonymous · Answer

For people on Plato looking back, Awanyulian was close but it's actually 89.03  g

thomaster · Answer

@WhatEven actually... it's 89.05478579