Informe de recreo

Resultados:
Método de Biuret
Datos obtenidos de Biuret |
Tubo | mL BSA | mL H2O | mL Biuret | %T | mg/mL BSA | Absorbancia(10-2) |
Blanco | 0,0 | 2,0 | 3,0 | 100 | 0,0 | 0,0 |
1 | 0,4 | 1,6| 3,0 | 95 | 1,0 | 2,2 |
2 | 0,8 | 1,2 | 3,0 | 81 | 2,0 | 9,2 |
3 | 1,2 | 0,8 | 3,0 | 70 | 3,0 | 15,5 |
4 | 1,6 | 0,4 | 3,0 | 48 | 4,0 | 31,9 |
5 | 2,0 | 0,0 | 3,0 | 43 | 5,0 | 36,7 |Muestra #3 | 1,3 | 0,7 | 3,0 | 61 | 3,25 | 21,5 |
Cálculos:
masa(mg) | Concentración(mg/mL) |
masa de la BSA=mL BSA×5,0mg BSA 1,0mL BSA | concentración=mg de BSAmL de muestra |
masa1 =0,4mL BSA×5,0mgBSA 1,0mL BSA=2,0mg | c1=2,0mg 2,0mL=1,0mg mL |
masa2 =0,8mL BSA ×5,0mgBSA 1,0mL BSA=4,0mg | c2=4,0mg 2,0mL=2,0mg mL |
masa3 =1,2mL BSA ×5,0mgBSA 1,0mL BSA=6,0mg | c3=6,0mg 2,0mL=3,0mg mL |masa4 =1,6mL BSA ×5,0mgBSA 1,0mL BSA=8,0mg | c4=8,0mg 2,0mL=4,0mg mL |
masa5 =2,0mL BSA ×5,0mgBSA 1,0mL BSA=10,0mg | c5=10,0mg 2,0mL=5,0mg mL |
masam3 =1,3mL BSA ×5,0mgBSA 1,0mL BSA=6,5mg |cm3=6,5mg 2,0mL=3,25mg mL |

Transmitancias(T=520nm) | Absorbancias(A=520nm) |
T=%T100% | A=-logT |
T1=%T1100%=95100=0,95 | A1=-logT1=-log0,95=2,2×10-2 |
T2=%T2100%=81100=0,81 |A2=-logT2=-log0,81=9,2×10-2 |
T3=%T3100%=70100=0,70 | A3=-logT3=-log0,70=15,5×10-2 |
T4=%T4100%=48100=0,48 | A4=-logT4=-log0,48=31,9×10-2 |
T5=%T5100%=43100=0,43 | A5=-logT5=-log0,43=36,7×10-2 |Tm3=%Tm3100%=61100=0,61 | Am3=-logTm3=-log0,61=21,5×10-2 |

% error=valorteorico-valorexperimentalvalorteorico×100%
% error=3,25mg/mL-3,26mg/mL3,25mg/mL×100%=0,01mg/mL3,25mg/mL×100%
% error≈0,31%Determinación de proteína bruta por el método Kjeldahl |
Muestra | Masa (g) | Volumen consumido (mL) | Factor de conversión |
Blanco | ------- | 0,8 | ------ |
Caseína | 0,2498 | 23,7 | 6,38 |Harina de pescado | 0,2505 | 19,1 | 6,25 |

Caseína
ml del titulante=ml del acido 0,1Nmuestra-ml del acido 0,1Nblanco
ml del titulante=23,7ml HCl -0,8ml HCl
ml del titulante=22,9ml HCl...