Physical and inorganic chemistry chapter 2 stoichiometry – Flashcards

6.022 x 10^23. example: 1 mol of C =6.022 x 10^23, 1 mol of HCl= 6.022 x 10^23, 2 mol of NaCl= 2 x 6.022 x 10^23 Number of Cl- ions in 2 moles of MgCl2 = 2 x 2 x 6.022 x 10^23 Number of Mg2+ ions in 1 mol of MgCl2 = 6.022 x 10^23

Atomic mass(neutron + proton) = Molar mass ( g/mol)

Number of moles= Mass of substance(Gram) / Molar mass of substance( g/mol)

Percent yield =(Actual product yield/ theoretical product yield )x 100%

%(w/v)= g/ml , %(v/v)= ml/ml , Molarity= mol/litre

Mi x Vi = Mf x Vi . Initial molarity x initial volume = final molarity x final volume.

pV = nRT . P= pressure, V= Volume of gas, n= number of mols of gas, R= gas constant, T= temperature of the gas (K). 1 atm=101325Pa =760mmHg .(1atm x 22.4L)/(1mol x 273K)=0.08206 (101325Pa x 0.0224m^3)/(1mol x 273K)= 8.314

mass percent of each element in a compound. Mass of copper in Cu2(OH)2CO3. [2x63.55/ (63.55 x 2 + (16x1.01) + 12.01 + 16x3)]x 100% = 57.5%

The simplest ratio of atoms in a compound. Mole ratio of Ba,P,O = 3:2:8 , Ba3P2O8

The actual number of atoms in a compound. molar mass of the compound is needed to dtermine the molecular formula. Molar mass of ( C3H4O3)n = 176 g/mol, (88.07)n= 176g/mol, n=2 Molecular formula= C6H8O6