Formulas for ionic compounds - 3 and sulfide ions(S 2...

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Formulas for ionic compounds empirical formula gives the elemental composition of a compound o formula lists elements present by element symbol o subscripts give ratios of ions or atoms in the compound CuSO 4 contains 4 atoms of O and 1 atom of S for every 1 atom of Cu. Na 2 CO 3 contains 2 atoms of Na and 3 atoms of O for every 1 atom of C. writing ionic empirical formulas o write the cation formulas, including charge o write the anion formulas, including charge o combine enough cations with enough anions to give a total charge of zero trick: swap charges as subscripts don't write charges when the ions are combined o use the simplest (lowest) cation-to-anion ratio possible o list cations first, anions last Potassium ions (K + ) and chloride ions (Cl - ) combine to give potassium chloride, KCl Calcium ions (Ca 2+ ) and bromide ions (Br - ) combine to give calcium bromide, CaBr 2 Aluminum ions (Al
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Unformatted text preview: 3+ ) and sulfide ions (S 2-) combine to give aluminum sulfide, Al 2 S 3 • naming ionic compounds from formulas o name the anions o name the cations recall that the names of transition metal and main group cations must include their charge as a Roman numeral. o the name of compound is cation name followed by anion name Na 2 S contains sodium ion and sulfide ion. The compound is sodium sulfide. SnCl 4 contains a tin cation and four chloride ions. Each chloride carries a -1 charge, so the tin must have a +4 charge. The compound is tin(IV) chloride. • the formula weight is the sum of atomic weights for atoms in the formula Na O H NaOH: 23.0 u + 16.0 u + 1.0 u 40.0 u Na C O Na 2 CO 3 2×23.0 u + 12.0 u + 3×16.0 u 106.0 u N H S O (NH 4 ) 2 SO 4 2× 14.0 u + 2×4×1.0 u + 32.0 u + 4×16.0 u 132.0 u...
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