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The number of formula units found in a 5.674 g sample of Mg(NO3)2Carbon

User Crazy Joe Malloy
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Step-by-step explanation:

We have to find the number of formula units or molecules found in 5.674 g of Mg(NO₃)₂. First we have to convert the grams into moles using the molar mass of Mg(NO₃)₂.

atomic mass of Mg = 24.31 amu

atomic mass of N = 14.01 amu

atomic mass of O = 16.00 amu

molar mass of Mg(NO₃)₂ = 24.31 + 2 * 14.01 + 6 * 16.00

molar mass of Mg(NO₃)₂ = 148.33 g/mol

moles of Mg(NO₃)₂ = mass of Mg(NO₃)₂/(molar mass of Mg(NO₃)₂)

moles of Mg(NO₃)₂ = 5.674 g/(148.33 g/mol)

moles of Mg(NO₃)₂ = 0.03825 moles

According to Avogadro's number there are 6.022 *10^23 formula units (or molecules) in 1 mol of Mg(NO₃)₂. We can use this relationship to find the answer to our problem.

1 mol of Mg(NO₃)₂ = 6.022 * 10^23 formula units

number of formula units = 0.03825 moles * 6.022 * 10^23 formula units/(1 mol)

number of formula units = 2.303 * 10^22 formula units

Step-by-step explanation: there are 2.303 * 10^22 formula units in 5.674 g of Mg(NO₃)₂

User Juergen Gutsch
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