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- Frage: (st-48)
Berechne die Molmasse folgender Verbindungen:
- Wasser
- Kohlendioxid
- C6H12O6
- NaCl
- Fe2O3
- CH4
- H2SO4
- C3H6O
Lösung: - M(H2O): 2·M(H) + 1·M(O) = 2·1+1·16 = 18 g/mol
- M(CO2): 1·M(C) + 2·M(O) = 1·12+2·16= 44 g/mol
- M(C6H12O6): 6·M(C) + 12·M(H) + 6·M(O) = 6+12+12·1+6·16 = 180 g/mol
- M(NaCl): M(Na+) + M(Cl–) = 23.0 +35.5 = 58.5 g/mol
- M(Fe2O3): 2·55.9+3·16=159.8 g/mol
- M(CH4) = 12+4·1 = 16 g/mol
- M(HSO4) = 2·1+324·16 = 98 g/mol
- M(C3H6O) = 3·12 6·1+16 = 58 g/mol
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