Plant Foods

Rajan Gill February 10, 2011

Experiment 4. Plant Foods

Abstract

Plant food was analysed and the amount of phosphorus present as phosphorus pentoxide in plant food, was determined through calculations. 1.5g of magnesium sulphate heptahydrate was dissolved in water and was then added to a solution containing 1g of plant food and water. Diluted aqueous ammonia was then added to the reaction mixture producing MgNH4PO4•H2O. The amount of phosphorus present in 1 gram of plant food was determined to be 8.03 x 10-2g.

Data and Results

Table 1. Preparation of MgNH4PO4•6H2O |

Chemical | Amount |

MgSO4•7H2O | 1.50 g |

Concentration of MgSO4•7H2O | 0.245 M |

Plant Food | 1.044 g |

Aqueous Ammonia | 5 mL |

Concentration of Ammonia | 5 M |

Ammonium Added | 0.075 mol |

isopropanol | 2 x 5 mL |

MgNH4PO4•6H2O | 0.6358g |

Table 2. Percent of Phosphate in P2O5 |

Chemical | Amount |

%P in MgNH4PO4•6H2O | 12.6% |

Mass of P in MgNH4PO4•6H2O | 8.03 x 10-2g |

%P in Plant Food | 7.69% |

%P present as P2O5 | 17.62% |

Advertised amount of P2O5 | 20.00% |

Percent of P recovered | 88.10% |

Calculations

1. Molarity of MgSO4.7H2O:

1.50g x (1 mol/246.38g) = 6.09 x 10-3 mol (1/0.025L) = 0.24M

Page 1

2. Moles of ammonium ions added:

15M = 15mol/L

15mol/L x (0.005L/1) = 0.075mol

3. Concentration of diluted ammonia solution:

V1 = 5mL

M2 = ?

V2 = 5 + 10 = 15mL

M1V1 = M2V2

(15M)(5mL) = (xM)(15mL)

x = 5M

4. % P in one mole of MgNH4PO4.6H2O:

MgNH4PO4.6H2O = 245.29g

P = 30.97g

%P = 30.97/245.29 = 12.6%

5. Mass of phosphorus in sample of MgNH4PO4.6H2O:

0.6358g x (30.97g/245.29g) = 8.03 x 10-2g

6. % of elemental phosphorus in plant food:

% P = (8.03 x 10-2g/1.044g) x 100 = 7.69%

7. % phosphorus present as phosphorus pentoxide:

P2O5 = 141.94g

% P2O5 = (7.69% x 141.94g)/(2 x30.97g) = 17.62%

8. % phosphorus...