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What's the amount of substance(mol) of Li+ in 0.3g of Li2CO3?

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We are given a sample of Li₂CO₃ (0.3 g) and we have to determine the number of moles of Li+ that we have in that sample. Using the molar mass of Li₂CO₃ we can calculate the number of moles of Li₂CO₃ that we have in 0.3 g of it, and then we will find the number of moles of Li+.

atomic mass of Li = 6.94 amu

atomic mass of C = 12.01 amu

atomic mass of O = 16.00 amu

molar mass of Li₂CO₃ = 2 * 6.94 + 1 * 12.01 + 3 * 16.00

molar mass of Li₂CO₃ = 73.89 g/mol

number of moles of Li₂CO₃ = mass of Li₂CO₃/(molar mass of Li₂CO₃)

number of moles of Li₂CO₃ = 0.3 g/(73.89 g/mol)

number of moles of Li₂CO₃ = 0.004 moles

Now in 1 molecule of Li₂CO₃ we have 2 ions Li+. So, in 1 mol of molecules of Li₂CO₃ we have 2 moles of ions Li+. We will use that relationship to find the amount of Li+ present in our sample.

number of moles of Li+ = 0.004 moles of Li₂CO₃ * 2 moles of Li+/(1 mol of Li₂CO₃)

number of moles of Li+ = 0.008 moles

Answer: the amount of substance of Li+ in 0.3 g of Li₂CO₃ is 0.008 moles

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