Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 3 - Mass Relationships in Chemical Reactions - Section Problems - Page 107: 56



Work Step by Step

$6.022 \times 10^{23} particles$ = $1 mol$ If it 25.00% is argon, we have $0.2500$ mol $Ar$ 1. Find the mass of argon (Molar mass = 39.95g/mol): $0.2500 mol \times \frac{39.95 g}{1 mol} = 9.987$ g 2. Calculate the mass of the other element: 25.12 g - 9.987 g = 15.13 g 3. If it is 75.00% the other element, then we have 0.7500 mol $X$. Calculate its molar mass: $\frac{15.13g}{0.7500mol} = 20.17$ g/mol. 4. Using a periodic table, we can see that the identity of the element with molar mass equal to $20.17$ is "Neon" (Ne).
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