An analysis of a 4.2-gram sample of a compound is found to contain 54.76 percent sodium (Na) and 45.24 percent fluorine (F). Determine the empirical formula of the compound. Na2F NaF2 NaF Na2F2

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An analysis of a 4.2-gram sample of a compound is found to contain 54.76 percent sodium (Na) and 45.24 percent fluorine (F). Determine the empirical formula of the compound. Na2F NaF2 NaF Na2F2

Chemistry
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If you think in terms of the hypothetical 100g sample, then you can calculate it like this: Number of moles (n) = mass (m) / molecular mass (Mr). You can find the molecular mass formula by looking in the periodic table. So; n(Na) = 54.76/22.99 = 2.381905... mol n(F) = 45.24/19.00 = 2.381052... mol Na has the smallest number of moles calculated and we will therefore divide each mole value on this smallest value, like this: Na: 2.381905... mol/2.381905... mol = 1 F: 2.381052... mol/2.381905... mol \[\approx 1\] So the empirical formula will then be: NaF

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